Worksheet on Enthalpy and Hess's Law with three problems for calculating ΔH using given chemical reactions and their enthalpy changes.

Here are the step-by-step solutions for the three problems on your worksheet.

Problem 1

Goal: Calculate $\Delta H$ for: $2 \text{CO}_2(g) + \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_2(g) + \frac{5}{2} \text{O}_2(g)$

Given Data:
1. $\text{C}_2\text{H}_2(g) + 2 \text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)$ ... $\Delta H = -94.5 \text{ kJ}$
2. $\text{H}_2\text{O}(g) \rightarrow \text{H}_2(g) + \frac{1}{2} \text{O}_2(g)$ ... $\Delta H = 71.2 \text{ kJ}$
3. $\text{C}_2\text{H}_6(g) + \frac{7}{2} \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + 3 \text{H}_2\text{O}(g)$ ... $\Delta H = -283 \text{ kJ}$

Step-by-Step Solution:
We need to rearrange these equations so they add up to our target equation.

* Step A: We need $\text{C}_2\text{H}_2$ on the product side (right). Equation 1 has it on the reactant side. Let's reverse Equation 1. When we reverse, we flip the sign of $\Delta H$.
* $\text{C}_2\text{H}_6(g) \rightarrow \text{C}_2\text{H}_2(g) + 2 \text{H}_2(g)$ ... $\Delta H = +94.5 \text{ kJ}$

* Step B: We need to get rid of the $\text{C}_2\text{H}_6$ we just created on the left. Equation 3 has $\text{C}_2\text{H}_6$ on the left. Let's reverse Equation 3 as well.
* $2 \text{CO}_2(g) + 3 \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_6(g) + \frac{7}{2} \text{O}_2(g)$ ... $\Delta H = +283 \text{ kJ}$

* Step C: Now let's look at what we have so far by adding Step A and Step B:
* Reactants: $\text{C}_2\text{H}_6 + 2 \text{CO}_2 + 3 \text{H}_2\text{O}$
* Products: $\text{C}_2\text{H}_2 + 2 \text{H}_2 + \text{C}_2\text{H}_6 + \frac{7}{2} \text{O}_2$
* Cancel $\text{C}_2\text{H}_6$: $2 \text{CO}_2 + 3 \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + 2 \text{H}_2 + \frac{7}{2} \text{O}_2$

* Step D: Compare this to our Target: $2 \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \frac{5}{2} \text{O}_2$.
* We have too much water ($3 \text{H}_2\text{O}$ instead of $1$). We need to remove $2 \text{H}_2\text{O}$ from the left.
* We have extra hydrogen gas ($2 \text{H}_2$) on the right that needs to go away.
* We have too much oxygen ($\frac{7}{2} \text{O}_2$ instead of $\frac{5}{2}$). We need to remove $1 \text{O}_2$ from the right.
* Look at Equation 2: $\text{H}_2\text{O} \rightarrow \text{H}_2 + \frac{1}{2} \text{O}_2$. If we reverse this, we get $\text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O}$. This removes $\text{H}_2$ and $\text{O}_2$ and adds $\text{H}_2\text{O}$.
* We need to do this twice to remove the $2 \text{H}_2$ and add $2 \text{H}_2\text{O}$ to balance things out. So, multiply reversed Equation 2 by 2.
* Reversed Eq 2: $\text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O}$ ... $\Delta H = -71.2 \text{ kJ}$
* Multiplied by 2: $2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$ ... $\Delta H = 2(-71.2) = -142.4 \text{ kJ}$

* Step E: Add all the modified enthalpies together:
1. Reverse Eq 1: $+94.5 \text{ kJ}$
2. Reverse Eq 3: $+283 \text{ kJ}$
3. Reverse & x2 Eq 2: $-142.4 \text{ kJ}$

Total $\Delta H = 94.5 + 283 - 142.4$
Total $\Delta H = 377.5 - 142.4$
Total $\Delta H = 235.1 \text{ kJ}$

***

Problem 2

Goal: Calculate $\Delta H$ for: $\text{Fe}(s) + \frac{3}{2} \text{Cl}_2(g) \rightarrow \text{FeCl}_3(s)$

Given Data:
1. $\text{Fe}(s) + \text{Cl}_2(g) \rightarrow \text{FeCl}_2(s)$ ... $\Delta H = -341.8 \text{ kJ}$
2. $\text{FeCl}_2(s) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{FeCl}_3(s)$ ... $\Delta H = -57.7 \text{ kJ}$

Step-by-Step Solution:
This one is simpler because the equations line up nicely if we just add them.

* Step A: Keep Equation 1 exactly as it is. It provides the starting Iron ($\text{Fe}$).
* $\text{Fe} + \text{Cl}_2 \rightarrow \text{FeCl}_2$ ... $\Delta H = -341.8 \text{ kJ}$

* Step B: Keep Equation 2 exactly as it is. It takes the product from Step A ($\text{FeCl}_2$) and turns it into the final product ($\text{FeCl}_3$).
* $\text{FeCl}_2 + \frac{1}{2} \text{Cl}_2 \rightarrow \text{FeCl}_3$ ... $\Delta H = -57.7 \text{ kJ}$

* Step C: Add the two equations together.
* Left Side: $\text{Fe} + \text{Cl}_2 + \text{FeCl}_2 + \frac{1}{2} \text{Cl}_2$
* Right Side: $\text{FeCl}_2 + \text{FeCl}_3$
* The intermediate compound $\text{FeCl}_2$ appears on both sides, so it cancels out.
* Combine the Chlorine gases: $1 \text{Cl}_2 + \frac{1}{2} \text{Cl}_2 = \frac{3}{2} \text{Cl}_2$.

Resulting Equation: $\text{Fe}(s) + \frac{3}{2} \text{Cl}_2(g) \rightarrow \text{FeCl}_3(s)$ (This matches the target!)

* Step D: Add the $\Delta H$ values.
* $\Delta H = -341.8 + (-57.7)$
* $\Delta H = -399.5 \text{ kJ}$

***

Problem 3

Goal: Calculate $\Delta H$ for: $2 \text{H}_2\text{O}_2(l) \rightarrow 2 \text{H}_2\text{O}(l) + \text{O}_2(g)$

Given Data:
1. $\text{H}_2(g) + \text{O}_2(g) \rightarrow \text{H}_2\text{O}_2(l)$ ... $\Delta H = -188 \text{ kJ/mol}$
2. $\text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)$ ... $\Delta H = -286 \text{ kJ/mol}$

Step-by-Step Solution:

* Step A: We need $2 \text{H}_2\text{O}_2$ on the reactant side (left). Equation 1 has $1 \text{H}_2\text{O}_2$ on the product side (right).
* First, reverse Equation 1: $\text{H}_2\text{O}_2(l) \rightarrow \text{H}_2(g) + \text{O}_2(g)$ ... $\Delta H = +188 \text{ kJ}$
* Second, multiply by 2 to get the coefficient correct: $2 \text{H}_2\text{O}_2(l) \rightarrow 2 \text{H}_2(g) + 2 \text{O}_2(g)$
* New $\Delta H = 2 \times (+188) = +376 \text{ kJ}$

* Step B: We need $2 \text{H}_2\text{O}$ on the product side (right). Equation 2 has $1 \text{H}_2\text{O}$ on the right.
* Multiply Equation 2 by 2: $2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(l)$
* New $\Delta H = 2 \times (-286) = -572 \text{ kJ}$

* Step C: Add the modified equations together.
* Eq A: $2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2 + 2 \text{O}_2$
* Eq B: $2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$
* Sum: $2 \text{H}_2\text{O}_2 + 2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2 + 2 \text{O}_2 + 2 \text{H}_2\text{O}$

* Step D: Simplify.
* Cancel $2 \text{H}_2$ from both sides.
* Cancel $1 \text{O}_2$ from the left with one of the $2 \text{O}_2$ on the right, leaving $1 \text{O}_2$ on the right.
* Final Equation: $2 \text{H}_2\text{O}_2(l) \rightarrow 2 \text{H}_2\text{O}(l) + \text{O}_2(g)$ (Matches target!)

* Step E: Add the $\Delta H$ values.
* $\Delta H = +376 + (-572)$
* $\Delta H = 376 - 572$
* $\Delta H = -196 \text{ kJ}$

Final Answer:
1. 235.1 kJ
2. -399.5 kJ
3. -196 kJ