Question

calculate the enthalpy for this reaction.
h₂o + c → co + h₂
use the following thermochemical equations.

1. h₂ + ½o₂ → h₂o δh₁ = −242.0 kj
2. 2co → 2c + o₂ δh₂ = 221.0 kj

δhᵣₓₙ = ? kj
enter either a + or - sign and the magnitude.
use significant figures.

Explanation:

Step1: Reverse Equation 1

Reverse the first equation \( \ce{H2 + \frac{1}{2}O2 -> H2O} \) to get \( \ce{H2O -> H2 + \frac{1}{2}O2} \). When we reverse a reaction, the sign of \( \Delta H \) changes. So \( \Delta H_{1}' = - \Delta H_{1} = -(-242.0\ \text{kJ}) = 242.0\ \text{kJ} \).

Step2: Reverse and Halve Equation 2

First, reverse the second equation \( \ce{2CO -> 2C + O2} \) to get \( \ce{2C + O2 -> 2CO} \), with \( \Delta H_{2}' = - \Delta H_{2} = -221.0\ \text{kJ} \). Then, halve this reversed equation to get \( \ce{C + \frac{1}{2}O2 -> CO} \). When we halve a reaction, we also halve the \( \Delta H \), so \( \Delta H_{2}'' = \frac{\Delta H_{2}'}{2} = \frac{-221.0\ \text{kJ}}{2} = -110.5\ \text{kJ} \).

Step3: Add the Modified Equations

Now, add the modified first equation (\( \ce{H2O -> H2 + \frac{1}{2}O2} \), \( \Delta H_{1}' = 242.0\ \text{kJ} \)) and the modified second equation (\( \ce{C + \frac{1}{2}O2 -> CO} \), \( \Delta H_{2}'' = -110.5\ \text{kJ} \)):
\[

$$\begin{align*} \ce{H2O -> H2 + \frac{1}{2}O2} &\quad \Delta H_{1}' = 242.0\ \text{kJ}\\ \ce{C + \frac{1}{2}O2 -> CO} &\quad \Delta H_{2}'' = -110.5\ \text{kJ}\\ \hline \ce{H2O + C -> CO + H2} &\quad \Delta H_{\text{rxn}} = \Delta H_{1}' + \Delta H_{2}'' \end{align*}$$

\]
Calculate \( \Delta H_{\text{rxn}} = 242.0\ \text{kJ} + (-110.5\ \text{kJ}) = 131.5\ \text{kJ} \).

Answer:

+131.5