Chemistry 120 Hess's Law Worksheet featuring five problems to calculate ΔH for different chemical reactions using provided thermochemical data.

To solve the problems using Hess's Law, we need to manipulate the given reactions and their enthalpy changes (ΔH) to match the target reaction. Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken. Here are the solutions for each problem:

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Problem 1:

Calculate ΔH for the reaction:
\[ \text{C}_2\text{H}_6(\text{g}) + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_8(\text{g}) \]

#### Given Reactions:
1. \( \text{C}_2\text{H}_6(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \quad \Delta H = -1411.0 \, \text{kJ/mol} \)
2. \( \text{C}_2\text{H}_6(\text{g}) + \frac{7}{2}\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l}) \quad \Delta H = -1560.0 \, \text{kJ/mol} \)
3. \( \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}) \quad \Delta H = -285.8 \, \text{kJ/mol} \)

#### Solution:
We need to find ΔH for:
\[ \text{C}_2\text{H}_6(\text{g}) + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_8(\text{g}) \]

First, let's write the target reaction in terms of the given reactions. Notice that the target reaction does not involve oxygen, so we need to eliminate oxygen from the given reactions.

1. From Reaction 1:
\[ \text{C}_2\text{H}_6(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \quad \Delta H_1 = -1411.0 \, \text{kJ/mol} \]

2. From Reaction 2:
\[ \text{C}_2\text{H}_6(\text{g}) + \frac{7}{2}\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l}) \quad \Delta H_2 = -1560.0 \, \text{kJ/mol} \]

3. From Reaction 3:
\[ \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}) \quad \Delta H_3 = -285.8 \, \text{kJ/mol} \]

To eliminate oxygen, we can subtract Reaction 1 from Reaction 2:
\[ \left( \text{C}_2\text{H}_6(\text{g}) + \frac{7}{2}\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l}) \right) \]
\[ - \left( \text{C}_2\text{H}_6(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \right) \]

This gives:
\[ \left( \frac{7}{2} - 3 \right)\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}) \]
\[ \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}) \quad \Delta H = \Delta H_2 - \Delta H_1 = -1560.0 - (-1411.0) = -149.0 \, \text{kJ/mol} \]

However, this is not directly useful. Instead, we can use Reaction 3 directly:
\[ \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}) \quad \Delta H_3 = -285.8 \, \text{kJ/mol} \]

Now, we need to combine these to form the target reaction. Notice that the target reaction is:
\[ \text{C}_2\text{H}_6(\text{g}) + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_8(\text{g}) \]

This can be achieved by:
\[ \text{C}_2\text{H}_6(\text{g}) + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_8(\text{g}) \]

Using the given data, we can see that:
\[ \Delta H = \Delta H_3 = -285.8 \, \text{kJ/mol} \]

Thus, the answer is:
\[ \boxed{-285.8} \]

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Problem 2:

Calculate ΔH for the reaction:
\[ 4\text{NH}_3(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 4\text{NO}(\text{g}) + 6\text{H}_2\text{O}(\text{g}) \]

#### Given Reactions:
1. \( \text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}(\text{g}) \quad \Delta H = -180.5 \, \text{kJ} \)
2. \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}) \quad \Delta H = -91.8 \, \text{kJ} \)
3. \( 2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{g}) \quad \Delta H = -483.6 \, \text{kJ} \)

#### Solution:
We need to find ΔH for:
\[ 4\text{NH}_3(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 4\text{NO}(\text{g}) + 6\text{H}_2\text{O}(\text{g}) \]

First, let's express the target reaction in terms of the given reactions. We need to manipulate the given reactions to match the target.

1. From Reaction 1:
\[ \text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}(\text{g}) \quad \Delta H_1 = -180.5 \, \text{kJ} \]

2. From Reaction 2:
\[ \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}) \quad \Delta H_2 = -91.8 \, \text{kJ} \]

3. From Reaction 3:
\[ 2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{g}) \quad \Delta H_3 = -483.6 \, \text{kJ} \]

We need to form 4 NH₃, 5 O₂, 4 NO, and 6 H₂O. Let's start by forming NH₃:
\[ 2 \times (\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})) \]
\[ 2\text{N}_2(\text{g}) + 6\text{H}_2(\text{g}) \rightarrow 4\text{NH}_3(\text{g}) \quad \Delta H = 2 \times (-91.8) = -183.6 \, \text{kJ} \]

Next, we need to form 4 NO:
\[ 2 \times (\text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}(\text{g})) \]
\[ 2\text{N}_2(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow 4\text{NO}(\text{g}) \quad \Delta H = 2 \times (-180.5) = -361.0 \, \text{kJ} \]

Finally, we need to form 6 H₂O:
\[ 3 \times (2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{g})) \]
\[ 6\text{H}_2(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 6\text{H}_2\text{O}(\text{g}) \quad \Delta H = 3 \times (-483.6) = -1450.8 \, \text{kJ} \]

Now, we combine these:
\[ 2\text{N}_2(\text{g}) + 6\text{H}_2(\text{g}) \rightarrow 4\text{NH}_3(\text{g}) \quad \Delta H = -183.6 \, \text{kJ} \]
\[ 2\text{N}_2(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow 4\text{NO}(\text{g}) \quad \Delta H = -361.0 \, \text{kJ} \]
\[ 6\text{H}_2(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 6\text{H}_2\text{O}(\text{g}) \quad \Delta H = -1450.8 \, \text{kJ} \]

Adding these:
\[ \Delta H = -183.6 + (-361.0) + (-1450.8) = -1995.4 \, \text{kJ} \]

Thus, the answer is:
\[ \boxed{-1995.4} \]

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Problem 3:

Calculate ΔH° for acetic acid, HC₂H₃O₂, using the following thermodynamic data:
\[ \text{HC}_2\text{H}_3\text{O}_2(\text{l}) + 2\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \quad \Delta H = -875.0 \, \text{kJ/mol} \]
\[ \text{C}(\text{s, graphite}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) \quad \Delta H = -394.51 \, \text{kJ/mol} \]
\[ \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}) \quad \Delta H = -285.8 \, \text{kJ/mol} \]

#### Solution:
We need to find the standard enthalpy of formation (ΔH°f) of acetic acid, HC₂H₃O₂(l). The standard enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its elements in their standard states.

The target reaction is:
\[ 2\text{C}(\text{s, graphite}) + 2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{HC}_2\text{H}_3\text{O}_2(\text{l}) \]

Using Hess's Law, we can manipulate the given reactions to match the target.

1. From Reaction 1:
\[ \text{HC}_2\text{H}_3\text{O}_2(\text{l}) + 2\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \quad \Delta H_1 = -875.0 \, \text{kJ/mol} \]

2. From Reaction 2:
\[ \text{C}(\text{s, graphite}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) \quad \Delta H_2 = -394.51 \, \text{kJ/mol} \]

3. From Reaction 3:
\[ \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l}) \quad \Delta H_3 = -285.8 \, \text{kJ/mol} \]

We need to reverse Reaction 1 to form HC₂H₃O₂:
\[ 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{HC}_2\text{H}_3\text{O}_2(\text{l}) + 2\text{O}_2(\text{g}) \quad \Delta H = +875.0 \, \text{kJ/mol} \]

Next, we need to form 2 CO₂ and 2 H₂O:
\[ 2 \times (\text{C}(\text{s, graphite}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g})) \]
\[ 2\text{C}(\text{s, graphite}) + 2\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) \quad \Delta H = 2 \times (-394.51) = -789.02 \, \text{kJ/mol} \]

\[ 2 \times (\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l})) \]
\[ 2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{l}) \quad \Delta H = 2 \times (-285.8) = -571.6 \, \text{kJ/mol} \]

Now, we combine these:
\[ 2\text{C}(\text{s, graphite}) + 2\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) \quad \Delta H = -789.02 \, \text{kJ/mol} \]
\[ 2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{l}) \quad \Delta H = -571.6 \, \text{kJ/mol} \]
\[ 2\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{HC}_2\text{H}_3\text{O}_2(\text{l}) + 2\text{O}_2(\text{g}) \quad \Delta H = +875.0 \, \text{kJ/mol} \]

Adding these:
\[ \Delta H = -789.02 + (-571.6) + 875.0 = -485.62 \, \text{kJ/mol} \]

Thus, the answer is:
\[ \boxed{-485.6} \]

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Problem 4:

Calculate ΔH for the reaction:
\[ \text{CH}_4(\text{g}) + \text{NH}_3(\text{g}) \rightarrow \text{HCN}(\text{g}) + 3\text{H}_2(\text{g}) \]

#### Given Reactions:
1. \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}) \quad \Delta H = -91.8 \, \text{kJ} \)
2. \( \text{C}(\text{s, graphite}) + 2\text{H}_2(\text{g}) \rightarrow \text{CH}_4(\text{g}) \quad \Delta H = -74.9 \, \text{kJ/mol} \)
3. \( \text{H}_2(\text{g}) + 2\text{C}(\text{s, graphite}) + \text{N}_2(\text{g}) \rightarrow 2\text{HCN}(\text{g}) \quad \Delta H = +270.3 \, \text{kJ} \)

#### Solution:
We need to find ΔH for:
\[ \text{CH}_4(\text{g}) + \text{NH}_3(\text{g}) \rightarrow \text{HCN}(\text{g}) + 3\text{H}_2(\text{g}) \]

Let's manipulate the given reactions to match the target.

1. From Reaction 1:
\[ \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}) \quad \Delta H_1 = -91.8 \, \text{kJ} \]

2. From Reaction 2:
\[ \text{C}(\text{s, graphite}) + 2\text{H}_2(\text{g}) \rightarrow \text{CH}_4(\text{g}) \quad \Delta H_2 = -74.9 \, \text{kJ/mol} \]

3. From Reaction 3:
\[ \text{H}_2(\text{g}) + 2\text{C}(\text{s, graphite}) + \text{N}_2(\text{g}) \rightarrow 2\text{HCN}(\text{g}) \quad \Delta H_3 = +270.3 \, \text{kJ} \]

We need to form CH₄, NH₃, HCN, and H₂. Let's start by reversing Reaction 2:
\[ \text{CH}_4(\text{g}) \rightarrow \text{C}(\text{s, graphite}) + 2\text{H}_2(\text{g}) \quad \Delta H = +74.9 \, \text{kJ/mol} \]

Next, we need to form NH₃:
\[ \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}) \quad \Delta H_1 = -91.8 \, \text{kJ} \]

Finally, we need to form HCN:
\[ \text{H}_2(\text{g}) + 2\text{C}(\text{s, graphite}) + \text{N}_2(\text{g}) \rightarrow 2\text{HCN}(\text{g}) \quad \Delta H_3 = +270.3 \, \text{kJ} \]

Combining these:
\[ \text{CH}_4(\text{g}) \rightarrow \text{C}(\text{s, graphite}) + 2\text{H}_2(\text{g}) \quad \Delta H = +74.9 \, \text{kJ/mol} \]
\[ \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}) \quad \Delta H = -91.8 \, \text{kJ} \]
\[ \text{H}_2(\text{g}) + 2\text{C}(\text{s, graphite}) + \text{N}_2(\text{g}) \rightarrow 2\text{HCN}(\text{g}) \quad \Delta H = +270.3 \, \text{kJ} \]

Adding these:
\[ \Delta H = +74.9 + (-91.8) + (+270.3) = +253.4 \, \text{kJ/mol} \]

Thus, the answer is:
\[ \boxed{+253.4} \]

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Problem 5:

Calculate ΔH for the reaction:
\[ 2\text{Al}(\text{s}) + 3\text{Cl}_2(\text{g}) \rightarrow 2\text{AlCl}_3(\text{s}) \]

#### Given Reactions:
1. \( 2\text{Al}(\text{s}) + 6\text{HCl}(\text{aq}) \rightarrow 2\text{AlCl}_3(\text{aq}) + 3\text{H}_2(\text{g}) \quad \Delta H = -1049.0 \, \text{kJ} \)
2. \( \text{HCl}(\text{g}) \rightarrow \text{HCl}(\text{aq}) \quad \Delta H = -74.8 \, \text{kJ/mol} \)
3. \( \text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{HCl}(\text{g}) \quad \Delta H = -1845.0 \, \text{kJ} \)
4. \( \text{AlCl}_3(\text{s}) \rightarrow \text{AlCl}_3(\text{aq}) \quad \Delta H = -323.0 \, \text{kJ/mol} \)

#### Solution:
We need to find ΔH for:
\[ 2\text{Al}(\text{s}) + 3\text{Cl}_2(\text{g}) \rightarrow 2\text{AlCl}_3(\text{s}) \]

Let's manipulate the given reactions to match the target.

1. From Reaction 1:
\[ 2\text{Al}(\text{s}) + 6\text{HCl}(\text{aq}) \rightarrow 2\text{AlCl}_3(\text{aq}) + 3\text{H}_2(\text{g}) \quad \Delta H_1 = -1049.0 \, \text{kJ} \]

2. From Reaction 2:
\[ \text{HCl}(\text{g}) \rightarrow \text{HCl}(\text{aq}) \quad \Delta H_2 = -74.8 \, \text{kJ/mol} \]

3. From Reaction 3:
\[ \text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{HCl}(\text{g}) \quad \Delta H_3 = -1845.0 \, \text{kJ} \]

4. From Reaction 4:
\[ \text{AlCl}_3(\text{s}) \rightarrow \text{AlCl}_3(\text{aq}) \quad \Delta H_4 = -323.0 \, \text{kJ/mol} \]

We need to form 2 AlCl₃(s). Let's start by reversing Reaction 4:
\[ \text{AlCl}_3(\text{aq}) \rightarrow \text{AlCl}_3(\text{s}) \quad \Delta H = +323.0 \, \text{kJ/mol} \]

Next, we need to form 6 HCl(aq):
\[ 6 \times (\text{HCl}(\text{g}) \rightarrow \text{HCl}(\text{aq})) \]
\[ 6\text{HCl}(\text{g}) \rightarrow 6\text{HCl}(\text{aq}) \quad \Delta H = 6 \times (-74.8) = -448.8 \, \text{kJ} \]

Finally, we need to form 3 Cl₂:
\[ 3 \times (\text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{HCl}(\text{g})) \]
\[ 3\text{H}_2(\text{g}) + 3\text{Cl}_2(\text{g}) \rightarrow 6\text{HCl}(\text{g}) \quad \Delta H = 3 \times (-1845.0) = -5535.0 \, \text{kJ} \]

Combining these:
\[ 2\text{Al}(\text{s}) + 6\text{HCl}(\text{aq}) \rightarrow 2\text{AlCl}_3(\text{aq}) + 3\text{H}_2(\text{g}) \quad \Delta H = -1049.0 \, \text{kJ} \]
\[ 6\text{HCl}(\text{g}) \rightarrow 6\text{HCl}(\text{aq}) \quad \Delta H = -448.8 \, \text{kJ} \]
\[ 3\text{H}_2(\text{g}) + 3\text{Cl}_2(\text{g}) \rightarrow 6\text{HCl}(\text{g}) \quad \Delta H = -5535.0 \, \text{kJ} \]
\[ 2\text{AlCl}_3(\text{aq}) \rightarrow 2\text{AlCl}_3(\text{s}) \quad \Delta H = 2 \times (+323.0) = +646.0 \, \text{kJ} \]

Adding these:
\[ \Delta H = -1049.0 + (-448.8) + (-5535.0) + 646.0 = -6386.8 \, \text{kJ} \]

Thus, the answer is:
\[ \boxed{-6386.8} \]

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Final Answers:

1. \(\boxed{-285.8}\)
2. \(\boxed{-1995.4}\)
3. \(\boxed{-485.6}\)
4. \(\boxed{+253.4}\)
5. \(\boxed{-6386.8}\)