1. A) 1L container with 5 mL of water held at 20°C
2. B) 2L container with 15 mL of water held at 20°C
3. C) 3L container with 500 mL of water held at 15°C
4. D) 2L container with 15 mL of water held at 25°C.

The 1L container with 5 mL held at 20°oC and the 2L container with 15mL of water held at 20°C are next with equal vapour pressure.

The 3L container with 500 mL held at 15°C has the lowest vapour pressure.

Here are the reasons:

- All containers contain the same liquid, so they all have the same intermolecular force of attraction in the liquid.

- The volume of liquid in the container does not affect vapour pressure.

- The volume of the container does not affect vapour pressure.

- Only the temperature affects vapour pressure. As temperature increases, vapour pressure also increases.  

In the given problem the only real variable is the temperature of the liquid. The highest temperature will yield the highest vapour pressure. The medium temperature will yield a slightly lesser vapour pressure. The low temperature will yield a third - and lowest - vapour pressure.

2NH₃(g) -> N₂(g) + 3H₂(g)

If the initial concentration of ammonia was 0.100M, calculate the equilibrium concentrations of each of the substances.

Thought process

1. The first thing we need to ask is what am I supposed to find?

We are asked to find the concentration of the substances at equilibrium.

2. We next need to ask what information we have.

We have the initial concentration of the ammonia, 0.100M.

We have the equilibrium constant for the dissociation of ammonia, K_c = 2.63 x 10^-9.

We also have a balanced equation with the reactants and the products. (If we did not have the equation we would have to write the equation first.)

3. We now need to decide how we can use the information that we have.

We can write the equation for the equilibrium constant of the reaction.

This equation links the equilibrium constant with the concentration of the reactants and products at equilibrium.

We have the equilibrium constant. We do not have any more information to put into this equation; we have 3 unknowns that we need to find.

Usually when we write an equation we try to find only one unknown. So we need to find a way to write the equation with one unknown to find.

We do this by preparing an Initial Change Equilibrium (ICE) Table.

How do we know what to put in the table?

We need:

1. An initial concentration of reactants and products.
2. An expression for the change in concentration of reactants and products.
3. An expression for the concentration of the reactants and products at equilibrium.

Initial Concentration 

We know that the initial concentration of ammonia is 0.100M. When we put the ammonia in the container, we did not put any hydrogen or nitrogen in the container. The initial concentration of hydrogen and nitrogen would both be 0.

Change in Concentration 

From the equation:

2NH₃(g) -> N₂(g) + 3H₂(g)

Let x be our single unknown concentration to be used in the equilibrium constant equation. So use the mole ratios of the reactants and products (2:1:3) from the equation of the reaction.

The amount of ammonia that reacted will be 2x. So the change in the concentration of ammonia will be –2x, because 2x amount of ammonia is no longer in the container.

Note: We usually try to keep the numbers in front of the substances from the equation as a part of our answer.

When 2x ammonia reacts, we have x amount of nitrogen and 3x amount of hydrogen being produced.

Equilibrium Concentration 

At equilibrium, x amount of nitrogen and 3x amount of hydrogen is present.

Since 2x amount of ammonia has reacted then 0.100 – 2x amount of ammonia is present at equilibrium.

The concentrations of ammonia, nitrogen and hydrogen at equilibrium can now be written with one unknown x. We can substitute the concentration of ammonia, nitrogen and hydrogen at equilibrium from the ICE table.

We can now place these into the equation, along with the equilibrium constant.

Find the square root of both sides of the equation:

We can use the quadratic formula to solve for x:

WE CAN NOW FIND THE CONCENTRATION OF EACH OF THE SUBSTANCES AT EQUILIBRIUM.

The concentration of nitrogen at equilibrium is x.

So the concentration of nitrogen at equilibrium is 0.000983M.

The concentration of hydrogen at equilibrium is 3x.
So the concentration of hydrogen at equilibrium is 3 x 0.000983M = 0.00295M.

The concentration of ammonia at equilibrium is 0.1 – 2x.
So the concentration of ammonia at equilibrium is 0.1 – 2(0.000983M) = 0.098M.

The reaction has an enthalpy of +155.7kJ.

How will the amount of product present at equilibrium be affected by the following?

Remember: figure out your equilibrium shift and then state how the product will be affected.

a. Adding (Increasing the concentration of the reactants)

According to Le Chatelier’s Principle if you increase the concentration of the reactant the reaction in equilibrium will try to remove the excess reactants. So the reaction will produce more products. The equilibrium will shift to the right to produce more products.
This increases the production of NO.

b. Removing (Decreasing the concentration of the reactants)

When the concentration of the reactants is decreased, the reaction in equilibrium will try to increase the amount of reactants. The product will break down to try to replace the reactant that was removed. The equilibrium will shift to the left to produce more reactants. This decreases the amount of product formed.

c. Adding a Catalyst

Adding a catalyst increases the rate of the reaction. This increases the rate of both the forward and the backward reaction equally.
Remember that it will make the time taken to achieve equilibrium faster, but it will not affect a system in equilibrium.

d. Increasing the Temperature

When the reaction is written as
This means that the forward reaction is endothermic and the backward reaction is exothermic. The forward reaction absorbs heat from the surroundings and the exothermic reaction adds heat to the surroundings.
When heat is increased, the system in equilibrium will try to remove the excess heat. It does this by increasing the endothermic reaction.

Since the forward reaction is endothermic, more products will be produced.

e. Increasing the pressure.

When pressure is increased, the system in equilibrium will try to remove the excess pressure. There are 2 moles of molecules on the left hand side of the reaction and 3 moles of molecules on the right hand side of the reaction.
More molecules present will increase the pressure.
Less molecules present will decrease the pressure.
When pressure is increased on a system in equilibrium, the system will try to decrease the pressure. It does this by trying to decrease the number of molecules.
The system will try to remove the excess pressure by decreasing the amount of products and shifting the equilibrium to the left.

f. Decreasing the Volume

From Boyle’s Law, volume is inversely proportional to pressure.
This means that when volume is increased, pressure is decreased and when volume is decreased, pressure is increased.
Therefore, the changes in volume will change the pressure. When volume is decreased, pressure is increased and the system will remove the excess pressure by decreasing the number of molecules.
The reaction will decrease the pressure by decreasing the amount of products. This is obtained by shifting the equilibrium to the left.

K_sp = 1.1 x 10^-10

1. Write a balanced equation for the reaction.

   BaCl_2(aq) + Na₂SO_4(aq) → BaSO_4(s) + 2NaCl_(aq)

2. Write an ionic equation for Barium Sulfate.

   Ba²⁺(aq) + SO₄^2-_(aq) ↔ BaSO_4(s)

3. Calculate the ion product. The equation for the ion product is:

   Q_sp = [Ba²⁺][SO₄^2-]

   Note that because BaSO₄ is a solid, you do not divide by the concentration of BaSO₄ in the ion product equation. To find the ion product, Q_sp you need to find the concentration of Barium and Sulfate ions in solution.

   Finding the concentration of Barium Ions

   Find moles of Ba²⁺ in solution, since Ba²⁺: BaSO₄ is 1:1, moles of BaSO₄ = Ba²⁺. Use molarity equation to solve for moles of initial solution. The concentration of Barium ions is the number of moles of Barium ions in 1000mL of solution. 4.0 x 10^-3M Barium Chloride means that 1000 mL solution contains 4.0 x 10^-3 moles of Barium ions.

   1 ml of the solution will contain 4.0 x 10^-3 / 1000 moles of Barium ions. 100mL of the solution will contain 100 x 4.0 x 10^-3 / 1000 moles of Barium ions, or 4 x 10^-4 moles of Barium ions.

4. Determine concentration of ions by dividing moles of Ba2+ by the total volume of both solutions that have now been mixed. Use molarity equation and divide moles by total volume of 400mL or 0.4L

   You added 100 mL and 300 mL of solution together so you have a total volume of 400 mL.

   One solution had 4 x 10^-4 moles of Barium ions, while the other solution had no Barium ions.

   As a result, 4 x 10^-4 moles of Barium ions is now in 400 mL of solution. Since we need the concentration of Barium ions present in the new solution, we need to find the number of moles of Barium ions that would be present in 1000 mL.

   400 mL solution has 4 x 10^-4 moles of Barium ions. 1mL solution has 4 x 10^-4/400 moles of Barium ions. 1000mL solution has 1000 x 4 x 10^-4/400 moles of Barium ions.

   The concentration of Barium ions is 0.001 M

   Finding the concentration of Sulfate ions

1. Find moles of SO4^2- in solution, since SO4^2- : Na₂SO₄ is 1:1, moles of Na₂SO₄ = SO4^2-. Use molarity equation to solve for moles of initial solution.

Step-by-step solution

The concentration of Sulfate ions is the number of moles of Sulfate ions in 100 mL of solution. 1000 mL Sodium Sulfate has 6.0 x 10^-4 moles Sulfate ions. 1mL solution of Sodium Sulfate has 6 x 10^-4/1000 moles of Sulfate ions. 300 mL of Sodium Sulfate solution has 300 x 6 x 10^-4/1000 moles of Sulfate ions, or 1.8 x 10^-4 moles of Sulfate ions.

2. Determine concentration of ions by dividing moles of SO₄4^2- by the total volume of both solutions that have now been mixed. Use molarity equation and divide moles by total volume of 400 mL or 0.4 L.

Step-by-step solution

You added 100 mL and 300 mL of solution together so you have a total volume of 400 mL.

One solution had 1.8 x 10^-4 moles of Sulfate ions, while the other solution had no Sulfate ions. As a result 1.8 x 10^-4 moles of Sulfate ions is now in 400 mL of solution

You need the concentration of the Sulfate ions, so you need to find out how many moles would be in 1000 mL of solution.

400 mL of the new solution has 1.8 x 10^-4 moles of Sulfate ions. 1 mL of the new solution has 1.8 x 10^-4/400 moles of Sulfate ions. 1000mL of the new solution has 1000 x 1.8 x 10^-4/400 moles of Sulfate ions.

The concentration of Sulfate ions is 4.5 x 10^-4

You now have the concentration of the Barium ions and the Sulfate ions in the solution.

Determine Q_sp

You can now place these values in the equation to find the ion product:

Q_sp = [Ba²⁺][SO₄^2-]

Q_sp = (0.001)(4.5 x 10^-4)

Q_sp = 4.5 x 10^-7

5. Compare the ion product Q^sp to the solubility product K^sp,

   Q_sp = 4.5 x 10^-7

   Recall that you were given K_sp = 1.1 x 10^-10

   Now Q_sp > K_sp

   This means that a Barium Sulfate precipitate will be formed.

Determine the pH of a 0.500M solution of acetic acid

Step 1

Write the chemical equation for dissolving acetic acid in water.

Step 2

Set up a table comparing the concentration of the acetic acid and the ions formed on the right hand side of the equation.

Step 3

Write the equation based on the acid constant.

Step 4

Now x is the concentration of both the and the ions.

Replace the concentrations in the equation with the concentrations at equilibrium and with the Ka value for acetic acid.

Step 5

This acid is weakly dissociated in water, so x is very small, and 0.5 – x is approximately equal to 0.5. Therefore you can rewrite this equation as

And solve this equation for x

 The concentration of the ion is 0.003M.

Step 6

The pH is the negative logarithm to the base 10 of the ion concentration.

Since you know the concentration of the ion you can put this value into the equation.

1. CH₄ is a tetrahedral molecule

Part 1: CH₄ is a tetrahedral molecule

The first thing to do with this sort of question is to identify the central atom. The central atom in this case is the carbon atom because all the hydrogen atoms are attached to the carbon atom.

This is shown in the Lewis dot-cross diagram below:

The next stepis to determine the number of electrons in the valence shell of the carbon atom. There are 4 valence electrons in the carbon atom and each of them is bonded to a hydrogen atom. There is a total of 8 electrons surrounding the central atom of the molecule:†† 4 pairs of electrons.

There are no lone pairs of electrons on the carbon atom.

This also indicates that in the CH₄ molecule, only the bonding pairs of electrons will determine the shape of the molecule.

Below is a Lewis Structure for methane that indicates this.

When drawn in three dimensions, this is the shape of methane:

The dashed lines represent a hydrogen that protrudes backwards. The dark triangle represents a hydrogen that protrudes forwards. (It is sticking outwards towards you.) The plain lines represent two hydrogens which are in the plane of the computer screen.

The furthest these bonds can be apart will be 109.5°.

This molecule is said to have a tetrahedral shape because if you imagine drawing lines joining only the atoms at the edge of the molecule (the hydrogen atoms) - the resulting shape is a tetrahedron.

Note carefully

When describing the shapes of molecules, lines joining only the atoms at the edges of the molecules are considered. These are not the lines representing the bonds. The lines representing the bonds are only drawn from the central atom to the other atoms. The lines representing the bonds are only used when the central atom is one of the atoms at the edge of the molecule.

1. What type of intermolecular forces of attraction exists between the molecules of the compounds?

The forces of attraction between these molecules are as follows:

Methane has dispersion forces because it is a non-polar molecule.
Hydrogen gas has dispersion forces because it is a covalently bonded molecule.
Methanol has hydrogen bonding because it is a polar molecule, with an oxygen-hydrogen bond.

2. How do these forces affect boiling point?

The methanol has hydrogen bonding between the molecules, so it has the highest boiling point.

Both methane and hydrogen gas has dispersion forces between the molecules.

3. How do I predict the difference between the boiling points of substances with the same forces of attraction between the molecules?

Methane (Molar Mass 16) is a larger molecule than hydrogen gas (Molar Mass 2), so it will have a larger boiling point.

Order of decreasing boiling points

Methanol – highest boiling point
Methane
Hydrogen gas – lowest boiling point.

Example 1

Write the name for the following structure.

Step 1: Identify the longest carbon chain and identify the parent name

There are 5 C in the longest chain, so the parent name is pentane (see table 1 below)

Step 2: Identify the functional group

You have –COOH as a functional group, so drop “e” from pentane and attach –oic acid (see table 3). You get pentanoic acid.

Step 3: Assign the position number

Step 4: Write the position number of the functional group if necessary.

Since the functional group is in the first C, you don’t have to write the position number.

Step 5: Identify the hydrocarbon branch.

There are three hydrocarbon branches, two (so need to write di-) and one . The one of the is attached to the third carbon and other to the fourth carbon. The is attached to the second carbon. Use table 1.

Therefore, the name of the structure is

Example 2

Write the structure for the following name.

3,4-diethyl-5-methyl-2-heptanol

Step 1: Write the number of carbons according to the parent name

Parent name is heptane, so there are 7 C in the longest carbon chain.

C – C – C – C – C – C – C

Step 2: Assign a number to each carbon of the parent (left to right)

Step 3: Write the functional group as indicated by the ending of the parent name at the appropriate position.

The ending is –ol and there is 2 in front of the parent name, so you need to write –OH at the 2nd carbon.

Step 4: Write any hydrocarbon branches at the appropriate position number, if any

You have ethyl at 3rd and 4th carbon, and methyl at 5th carbon.

Step 5/step 6:

Complete the structure by adding the appropriate numbers of hydrogen to each carbon. Carbon must have four bonds. Delete the position number.

1. One 2s orbital and one 2p orbital are mixed?
2. One 2s orbital and two 2p orbitals are mixed?
3. One 2s orbital and three 2p orbitals are mixed?
4. One 2s orbital, three 3p orbitals and one 3d orbital?
5. One 3s orbital, three 3p orbitals and two 3d orbitals?

Part 1: One 2s orbital and one 2p orbital are mixed.

When one 2s and one 2p orbital are mixed, the sp hybrid orbitals are produced. The shapes of the orbitals are demonstrated below.

There are two atomic orbitals combined, so two hybrid orbitals are produced.

The diagram below represents a hybrid orbital.

The following diagram represents how the two sp hybrid orbitals are arranged on an atom. The orbitals are represented in different colours just to highlight the different hybrid orbitals.

Part 2: One 2s orbital and two 2p orbitals are mixed.

When one 2s orbital and two 2p orbitals are mixed, the hybrid orbitals are called sp² hybrid orbitals. The superscript 2 in sp² shows the number of p orbitals that were used in formation of the hybrid orbitals. When there is only one orbital there is no subscript.

For example, there is only one s orbital in the hybrid orbital, so there is no 1 written as a subscript. There are three original atomic orbitals (one s and two p orbitals) and three hybrid orbitals.

One way to check the number of hybrid orbitals formed is to write on your rough work paper the s¹p².

Remember that although no 1 is written as a subscript in the sp² hybrid orbitals, there is one s orbital. Then and add the superscripts 1 + 2 = 3.

There are three sp² hybrid orbitals.

The following diagram represents how the three sp² hybrid orbitals are arranged on an atom. Each sp² hybrid orbital is represented in a different colour.

Three sp² hybrid orbitals

Part 3: One 2s orbital and three 2p orbitals are mixed.

When one 2s and three 2p orbitals are mixed, the hybrid orbital is called sp³ hybrid orbitals. There are four hybrid orbitals formed from a total of four atomic orbitals (one s and three p).

In order to check the number of hybrid orbitals formed, write on your rough work paper the s¹p³.

Then add the superscripts 1 + 3 = 4.

There are four sp³ hybrid orbitals.

The following diagram represents how the four sp³ hybrid orbitals are arranged on an atom. Each sp³ hybrid orbital is represented in a different colour.

Four sp³ hybrid orbitals

Part 4: One 3s orbital, three 3p orbitals and one 3d orbital.

When one 3s, three 3p and one 3d orbital are mixed, the type of hybrid orbitals formed are called dsp³ orbitals. There are five original atomic orbitals (one s, three p and one d), so there are five dsp³ hybrid orbitals produced.

One 3s + three 3p + one 3d "→" five dsp³ hybrid orbitals.

The following is a representation of the five dsp³ hybridized orbitals on one atom.

Five dsp³ hybrid orbitals

The way to check the number of hybrid orbitals formed is to write on your rough work paper the d¹s¹p³. Then add the superscripts 1 + 1 + 3 = 5.

There are five dsp³ hybrid orbitals.

Part 5: One 3s orbital, three 3p orbitals and two 3d orbitals.

When one 3s, three 3p and two 3d orbital are mixed, the type of orbitals formed are called d²sp³ orbitals. There are six original atomic orbitals (one s, three p and two d), so there are six d²sp³ hybrid orbitals produced.

One 3s + three 3p + two 3d "→" six d²sp³ hybrid orbitals.

The following is a representation of the six d²sp³ hybridized orbitals on an atom.

Six d²sp³ hybrid orbitals

You can check the number of hybrid orbitals formed by writing on your rough work paper the d²s¹p³. Then add the superscripts 2 + 1 + 3 = 6.

There are six d²sp³ hybrid orbitals.

Example 2

2KCl + MnO₂ + 2H₂SO₄ → K₂SO₄ + MnSO₄ + Cl₂ + 2H₂O

To answer this problem, you will apply the following steps:

1. Determine the oxidation number for every atom in the reaction

   2KCl + MnO₂ + 2H₂SO₄ → K₂SO₄ + MnSO₄ + Cl₂ + 2H₂O

   Reactants:
    

   2KCl	MnO2	2H2SO4
   K = +1	O = -2, but there are 2 of them Therefore, (-2) x 2 = -4	O = -2, but there are 4 of them Therefore, (-2) x 2 = -8
   Cl = -1	To find Mn; use the following equation Mn + (-4) = 0 (because the net charge of the substance is zero) Therefore, Mn = -4	H = +1, but there are 2 of them Therefore, (+1) x 2 = +2
   		To solve for S; use the following equation (+2) + S + (-8) = 0 S - 6 = 0 S = +6

   Products:
    

   K2SO4	MnSO4	Cl2	2H2O
   O = -2, but there are 4 of them Therefore, (-2) x 2 = -8	O = -2, but there are 4 of them Therefore, (-2) x 2 = -8	Since chlorine is in its elemental state, its oxidation number is zero Cl = 0	O = -2
   K = +1, but there are 2 of them Therefore, (+1) x 2 = +2	Since S is part of the sulphate ion, its oxidation number stays the same S = +6		H = +1, and there are two of them
   To solve for S; use the following equation (+2) + S + (-8) = 0 S - 6 = 0 S = +6	To solve for Mn; use the following equation Mn + (+6) + (-8) =0 Mn + (-2) = 0 Mn = +2

   ** NOTE: The oxidation number is the value before the number of atoms is multiplied to it. Therefore, oxygen’s oxidation number is -2. We multiply the oxidation number by the number of atoms to determine other atom’s oxidation numbers.

2. Write the 2 half-reactions
   1. Take the 1^st atom that has an oxidation number that is changing and write only it and the compound it is part of, in an equation.

      Chlorine has changed from -1 to zero

      2KCl → Cl₂
      (-1) → (0)

   2. Take the 2^nd atom that has an oxidation number that is changing and write only it and the compound it is part of, in an equation.

      Manganese has changed from +4 to +2

      MnO₂ → MnSO₄
      (+4) → (+2)

3. Add in the appropriate number of electrons on either the reactant side if they have gained electrons, or the product side if they lost electrons.
    

   2KCl → Cl₂ + 2e^-

   MnO₂ + 2e^- → MnSO₄

4. Determine which substance is oxidized or reduced:
   
5. Whichever substance gained electrons will be the reduced substance, or the oxidizing agent. Whichever substance lost electrons will be the oxidized substance, or the reducing agent. 2KCl → Cl2 + 2e-
6. 
   - lost electrons; therefore, oxidized and is the reducing agent MnO2 + 2e- → MnSO4
   - gained electrons; therefore, reduced and is the oxidizing agent

Thought process

1. Because this equation is in a basic medium, we follow example c. Ion-electron method in a basic medium.
    

2. You must follow the 10 steps in order to ensure that you get the correct answer to the question.
    

3. Now, go through each step in the process.

1. - A barium nitrate solution reacts with a sodium sulphate solution.

Step 1
Write a balanced chemical equation indicating the state of each species. You will need to use your solubility chart for this.

Ba(NO₃)_2(aq)+ Na₂SO_4(aq) → BaSO_4(s) + 2NaNO_3(aq)

Step 2
Rewrite the equation, separating the ions that are in solution. At this stage you have written the complete ionic equation.

Since all the reactants are aqueous solutions, this means that the ions are able to move freely. Therefore, you can separate the compounds into their ions.

Ba²⁺_(aq) + 2NO₃^-_(aq) +2Na⁺_(aq) +SO₄^2-_(aq) → BaSO_4(s) + 2Na⁺_(aq)+2NO₃^-_(aq)

Note that (NO₃)₂ from the Ba(NO₃)₂ is written as 2NO₃^- because the ions are now separated, and for each barium nitrate in solution, two nitate ions and one barium ion are produced.

The same applies to all of the 2Na⁺ ions.

Note that solids and pure liquids do not separate into ions.

Step 3
Identify and cross out spectator ions.
Spectator ions are ions that are exactly the same on both sides of the reaction. They are ions that have the same oxidation state and physical state on both sides of the equation.

The ions that are the same on both sides of the equation are 2Na+(aq) and 2NO3-(aq) Therefore, these are the spectator ions.

Ba²⁺_(aq)+ 2NO₃^-_(aq) + 2Na⁺_(aq) +SO₄^2-_(aq) → BaSO_4(s) + 2Na⁺_(aq) +2NO₃^-_(aq)

Step 4
Write the other ions that are left and you will have the net ionic equation.

Ba²⁺_(aq) + SO₄^2-_(aq) → BaSO_4(s)

Note that the ions left have changed physical states.
They were in aqueous solutions and they have combined to produce a solid.

How do I determine the balanced chemical equation, complete ionic equation and net ionic equation for a specific reaction?

Thought process

This question is asking you to find the change in enthalpy for the decomposition reaction.

Notice that you are only given enough information to write the final reaction’s equation, and you are not given a series of intermediate reactions.

Therefore, you will use the Hess’s Law Equation to solve this problem.

Step 1:

Write the balanced chemical equation for the reaction

2NaHCO₃ (s) → Na₂CO₃(s) + H₂O(l) + CO₂(g)

Step 2:

Write the Hess’s Law Equation

ΔΗ^o_reaction= (ΣΔΗ^o_{f products}) - (ΣΔΗ^o_{f reactants})

ΔΗ^o_reaction = {(1 mol Na₂CO₃(s) x ΔΗ^o_f Na₂CO₃(s) ) + (1 mol H₂O(l) x ΔΗ^o_f H₂O(l)) +

(1mol CO₂(g) x ΔΗ^o_f CO₂(g))} - {(2 mol NaHCO₃ (s) x ΔΗ^o_f NaHCO₃(g))}

Step 3:

Substitute in the values, found in the Chart of Thermodynamic Values (usually found in the appendix of your text book) and the appropriate coefficient.

ΔΗ^o_reaction = {(1 mol Na₂CO₃(s) x -1131kJ/mol ) + (1 mol H₂O(l) x -285.9kJ/mol) +

(1 mol CO₂(g) x -393.5kJ/mol)} - {(2 mol NaHCO₃ (s) x -947.7kJ/mol)}

Step 4:

Solve the equation

ΔΗ^o_reaction = (-1810 kJ) –(-1895kJ)

ΔΗ^o_reaction = +85kJ

Since the answer is positive, the reaction is endothermic.

Remember in doing these problems

1. Your final units should be in kilojoules (kJ).
2. If your answer is positive, the reaction is endothermic.
3. If your answer is negative, the reaction is exothermic.
4. Different compounds, and different states will be in your reaction.
6. Take note of the coefficients in your balanced chemical equation. These will change depending on your reaction and affect the number of moles in the enthalpy multiplication.

Thought process

This question is asking you to find the change in entropy for a decomposition reaction. You should use the standard entropy change equation to solve this problem.

The following steps will guide you through to the answer

Step 1:

Write the balanced chemical equation for the reaction

2NaHCO₃(s) → Na₂CO₃(s) + H₂O(l) + CO₂(g)

Step 2:

Write the Standard Entropy Change Equation and the appropriate coefficient

ΔS^o_reaction = (ΣΔS^o _products) - (ΣΔS^o _reactants)

ΔS^o_reaction = {(1 mol Na₂CO₃(s) x ΔS^o Na₂CO₃(s)) + (1 mol H₂O(l) x ΔS^o H₂O(l)) + (1 mol CO₂(g) x ΔS^o CO₂(g))} - {(2 mol NaHCO₃ (s) x ΔS^o NaHCO₃(s))}

Step 3:

Substitute in the values, found in the Chart of Thermodynamic Values. (Look in the appendix of your text book.)

ΔS^o_reaction = {(1 mol Na₂CO₃(s) x +136 J/molK ) + (1 mol H₂O(l) x +69.96 J/molK) +

(1 mol CO₂(g) x +213.6J/molK)} -{(2 mol NaHCO₃ (s) x +102 J/molK)}

Notice that the moles cancel out, leaving your units as J/K

Step 4:

Solve the equation

ΔS^o_reaction = +419.56J/K - (+204J/K)

ΔS^o_reaction = +215.56 J/K

Since the answer is positive, the reaction favours spontaneity.

Remember that your final units should be in Joules/Kelvin.

Also, if your answer is positive, the reaction favours spontaneity.

A sample of copper was heated to and then thrust into 200 g of water at The temperature of the mixture became

a) How much heat in joules was absorbed by the water?
b) How much heat in joules was lost by the copper?
c) What was the mass in grams of the copper sample?

Draw a picture:

Since the temperature of the copper is greater than the water, copper will release energy while water absorbs.

a. You were asked to find the heat energy, so you would use

b. You know that q (absorbed) = - q(released), so

q(copper) = - 1379.4 J

c. You found q(copper) = - 1379.4 J from part (b), and you also know that , so

Thought process

1. In this example, the question is giving you the starting and finishing temperature of the pure substance, water. It has also told you that you are starting with a liquid and are finishing with a solid. Therefore, you know the main parts of the graph.
3. The question is also asking you to solve for the mass. Therefore, you need to use the Heat Capacity Equation to determine this value. Remember that the equation is:
4. 
   
6. The second part of the question asks you to sketch the cooling curve for the reaction.
   
   Cooling curve:
7. 1. The water is being cooled, thus the heat energy is being taken away and the temperature begins to drop. Therefore, the kinetic energy begins to decrease and the molecules begin to lose speed, and the amount of space between molecules decreases. As the space between molecules decreases, the intermolecular force of attraction begins to increase. This decrease in temperature, speed and distance continues until the force of attraction between the molecules is strong enough to change the intermolecular bonds and create a solid. At this point, the molecules are "solidifying" and changing state from a liquid to a solid.
      Now the temperature of the gas will no longer decrease, because the heat energy is being used to form the intermolecular bonds between the liquid molecules. This plateau is the change of state from gas to liquid known as condensation.
      
      Cooling Curve
      
      
      
       
   2. Once all of the liquid molecules have changed into solid molecules, the heat energy will continue to be removed from the solid molecules causing them to lose kinetic energy and to decrease their temperature. This loss in kinetic energy will cause the molecules to move even more slowly and to come together even more and their temperature will continue to decrease.
      
      Cooling Curve
      
      
      
       
   3. Once all of the liquid molecules have changed into solid molecules, the heat energy will be removed from the solid molecules causing them to lose kinetic energy and to decrease their temperature.
       
   4. The above sketch is the graph of the cooling process.