Question

thermochemistry workshop
in problems 1 - 3, calculate δh for the reaction:

1. using the thermochemical equations (hesss law)

c₂h₅oh(ℓ) + 3 o₂(g) → 2 co₂(g) + 3 h₂o(g)
δh = - 1234.7 kj
c₂h₄(g) + 3 o₂(g) → 2 co₂(g) + 2 h₂o(g)
δh = - 1322.9 kj
2 co₂(g) + 3 h₂o(g) → c₂h₅oh + 3 o₂(g)
δh = - 88.2 kj

Explanation:

Step1: Reverse the first equation

$$\mathrm{C_2H_4(g) + 3O_2(g) ightarrow 2CO_2(g) + 2H_2O(g)}\quad \Delta H=-1322.9\ \text{kJ}$$

Reversed:

$$\mathrm{2CO_2(g) + 2H_2O(g) ightarrow C_2H_4(g) + 3O_2(g)}\quad \Delta H = 1322.9\ \text{kJ}$$

Step2: Use Hess's Law

We want

$$\mathrm{C_2H_4(g) + H_2O(g) ightarrow C_2H_5OH(l)}$$

Let the second equation be

$$\mathrm{C_2H_5OH(l)+3O_2(g) ightarrow 2CO_2(g) + 3H_2O(g)}\quad \Delta H=-1234.7\ \text{kJ}$$

Add the reversed first equation and the second equation:

$$\mathrm{(2CO_2(g) + 2H_2O(g) ightarrow C_2H_4(g) + 3O_2(g))}+$$

$$\mathrm{(C_2H_5OH(l)+3O_2(g) ightarrow 2CO_2(g) + 3H_2O(g))}$$

Cancel out common terms ($\mathrm{2CO_2(g)}$, $\mathrm{3O_2(g)}$). We get

$$\mathrm{C_2H_4(g) + H_2O(g) ightarrow C_2H_5OH(l)}$$

For $\Delta H$: $\Delta H=(1322.9)+(- 1234.7)=88.2\ \text{kJ}$

Answer:

$\Delta H=-88.2\ \text{kJ}$