Question

consider the following intermediate chemical equations.
c(s)+o₂(g)→co₂(g) δh₁=-393.5kj
2co(g)+o₂(g)→2co₂(g) δh₂=-566.0kj
2h₂o(g)→2h₂(g)+o₂(g) δh₃=483.6kj
the overall chemical equation is c(s)+h₂o(g)→co(g)+h₂(g). to calculate the final enthalpy of the overall chemical equation, which step must occur?

• reverse the first equation, and change the sign of the enthalpy. then, add.
• reverse the second equation, and change the sign of the enthalpy. then, add.
• multiply the first equation by three, and triple the enthalpy. then, add.
• divide the third equation by two, and double the enthalpy. then, add.

Explanation:

Step1: Recall Hess's Law

Hess's Law states that the enthalpy change of a reaction is the same regardless of the path taken, so we can manipulate intermediate equations (reverse, multiply/divide) to get the overall equation, adjusting enthalpy accordingly.

Step2: Analyze each option

- **Option 1 (Reverse first equation):** The first equation is \( C(s) + O_2(g) ightarrow CO_2(g) \). Reversing it would give \( CO_2(g) ightarrow C(s) + O_2(g) \), which doesn't help form the overall equation. Eliminate.
- **Option 2 (Reverse second equation):** The second equation is \( 2CO(g) + O_2(g) ightarrow 2CO_2(g) \) with \( \Delta H_2 = -566.0 \, \text{kJ} \). Reversing it gives \( 2CO_2(g) ightarrow 2CO(g) + O_2(g) \), and we change the sign of \( \Delta H_2 \) to \( +566.0 \, \text{kJ} \). Now, let's combine with other equations:
- First equation: \( C(s) + O_2(g) ightarrow CO_2(g) \) (\( \Delta H_1 = -393.5 \, \text{kJ} \))
- Reversed second equation: \( 2CO_2(g) ightarrow 2CO(g) + O_2(g) \) (\( \Delta H_2' = +566.0 \, \text{kJ} \))
- Third equation: \( 2H_2O(g) ightarrow 2H_2(g) + O_2(g) \) (\( \Delta H_3 = 483.6 \, \text{kJ} \)) – wait, no, we need to adjust the third equation too? Wait, let's check the overall equation: \( C(s) + H_2O(g) ightarrow CO(g) + H_2(g) \). Let's manipulate equations:
1. First equation: \( C(s) + O_2(g) ightarrow CO_2(g) \) (\( \Delta H_1 \))
2. Reverse second equation: \( 2CO_2(g) ightarrow 2CO(g) + O_2(g) \) (\( -\Delta H_2 \))
3. Divide third equation by 2: \( H_2O(g) ightarrow H_2(g) + \frac{1}{2}O_2(g) \) (\( \frac{\Delta H_3}{2} \))
Now, add equation 1, reversed equation 2 (divided by 2? Wait, no, reversed second equation is \( 2CO_2 ightarrow 2CO + O_2 \), so if we take equation 1 (1 \( CO_2 \) produced), reversed equation 2 (2 \( CO_2 \) consumed) – no, maybe better to scale. Wait, let's do the math:
Overall equation: \( C(s) + H_2O(g) ightarrow CO(g) + H_2(g) \). Let's combine:
- Equation 1: \( C + O_2 ightarrow CO_2 \)
- Reversed equation 2 (divided by 2? Wait, no, reversed second equation is \( 2CO_2 ightarrow 2CO + O_2 \), so if we take equation 1 (1 \( CO_2 \)), reversed equation 2 (2 \( CO_2 \)) – no, let's adjust coefficients. Wait, maybe I made a mistake. Wait, the overall equation has 1 \( CO \), so reversed second equation (2 \( CO \)) – so we need to divide reversed second equation by 2? Wait, no, the option says "reverse the second equation, and change the sign of the enthalpy. Then, add." Let's check the coefficients:
After reversing the second equation, we have \( 2CO_2 ightarrow 2CO + O_2 \). Then, add the first equation: \( C + O_2 ightarrow CO_2 \). Now, we have \( C + O_2 + 2CO_2 ightarrow CO_2 + 2CO + O_2 \). Simplify: \( C + CO_2 ightarrow 2CO \). Not helpful. Wait, maybe I messed up. Wait, the correct approach:
Let's write the target equation: \( C(s) + H_2O(g) ightarrow CO(g) + H_2(g) \). Let's express it as a combination:
1. \( C(s) + O_2(g) ightarrow CO_2(g) \) (\( \Delta H_1 \))
2. \( 2CO_2(g) ightarrow 2CO(g) + O_2(g) \) (reverse of second equation, \( -\Delta H_2 \))
3. \( 2H_2O(g) ightarrow 2H_2(g) + O_2(g) \) (\( \Delta H_3 \)) – no, we need \( H_2O(g) \), so divide equation 3 by 2: \( H_2O(g) ightarrow H_2(g) + \frac{1}{2}O_2(g) \) (\( \frac{\Delta H_3}{2} \))
Now, add equation 1, reversed equation 2 (divided by 2? Wait, reversed equation 2 is \( 2CO_2 ightarrow 2CO + O_2 \), so if we take equation 1 (1 \( CO_2 \)), reversed equation 2 (2 \( CO_2 \)) – no, scale equation 1 by 1, reversed equation 2 by 1/2: \( CO_2 ightarrow CO + \frac{1}{2}O_2 \) (\( -\frac{\Delta H_2}{2} \)). Then:
Equation 1: \( C + O_2 ightarrow CO_2 \) (\( \Delta H_1 \))
Reversed equation 2 (1/2): \( CO_2 ightarrow CO + \frac{1}{2}O_2 \) (\( -\frac{\Delta H_2}{2} \))
Equation 3 (1/2): \( H_2O ightarrow H_2 + \frac{1}{2}O_2 \) (\( \frac{\Delta H_3}{2} \))
Now, add them: \( C + O_2 + CO_2 + H_2O ightarrow CO_2 + CO + \frac{1}{2}O_2 + H_2 + \frac{1}{2}O_2 \). Simplify: \( C + H_2O ightarrow CO + H_2 \), which is the overall equation! Wait, but the option says "reverse the second equation, and change the sign of the enthalpy. Then, add." Wait, maybe I overcomplicated. Let's check other options:
- **Option 3 (Multiply first equation by 3):** Multiplying the first equation by 3 would give \( 3C + 3O_2 ightarrow 3CO_2 \), which is irrelevant to the overall equation. Eliminate.
- **Option 4 (Divide third equation by 2, double enthalpy):** Dividing the third equation by 2 gives \( H_2O(g) ightarrow H_2(g) + \frac{1}{2}O_2(g) \) with \( \Delta H = \frac{483.6}{2} = 241.8 \, \text{kJ} \). Doubling the enthalpy would undo the division, so this is contradictory. Eliminate.

Wait, maybe my initial analysis of Option 2 was correct. Let's re-express the steps:
1. First equation: \( C + O_2 ightarrow CO_2 \) (\( \Delta H_1 \))
2. Reverse second equation: \( 2CO_2 ightarrow 2CO + O_2 \) (\( -\Delta H_2 \))
3. Divide third equation by 2: \( H_2O ightarrow H_2 + \frac{1}{2}O_2 \) (\( \frac{\Delta H_3}{2} \))
Now, add equation 1, reversed equation 2, and divided third equation:
- Equation 1: \( C + O_2 ightarrow CO_2 \)
- Reversed equation 2: \( 2CO_2 ightarrow 2CO + O_2 \)
- Divided third equation: \( H_2O ightarrow H_2 + \frac{1}{2}O_2 \)
Wait, the \( O_2 \) from equation 1 and reversed equation 2: \( O_2 \) (from 1) + \( O_2 \) (from reversed 2) – no, equation 1 produces \( CO_2 \), reversed equation 2 consumes \( 2CO_2 \), so we need to scale equation 1 by 2? Wait, no, the overall equation has 1 \( C \), so equation 1 is 1 \( C \). Reversed equation 2 is 2 \( CO_2 \), so we need 2 \( CO_2 \) from equation 1? No, equation 1 gives 1 \( CO_2 \) per \( C \). This is getting confusing, but the key is that reversing the second equation (and changing its enthalpy sign) is necessary to introduce \( CO(g) \) in the products, which matches the overall equation. The other options are clearly incorrect (multiplying first by 3, dividing third and doubling enthalpy is illogical, reversing first doesn't help). So Option 2 is correct.

Answer:

B. Reverse the second equation, and change the sign of the enthalpy. Then, add.