To find the enthalpy change for the reaction [tex]\( \text{CO} + 2\text{H}_2 \rightarrow \text{CH}_3\text{OH} \)[/tex], we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for any set of reactions that add up to the overall reaction. Here are the steps to find the enthalpy change:
1. Identify the given thermochemical equations and their enthalpies:
- Equation 1: [tex]\(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\)[/tex], [tex]\(\Delta H_1 = -571 \text{kJ}\)[/tex]
- Equation 2: [tex]\(2\text{CO} + \text{O}_2 \rightarrow 2\text{CO}_2\)[/tex], [tex]\(\Delta H_2 = -566 \text{kJ}\)[/tex]
- Equation 3: [tex]\(2\text{CH}_3\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 4\text{H}_2\text{O}\)[/tex], [tex]\(\Delta H_3 = -1430 \text{kJ}\)[/tex]
2. Manipulate the given equations to derive the target reaction:
- First, start with the reverse of the third equation:
[tex]\[
2\text{CO}_2 + 4\text{H}_2\text{O} \rightarrow 2\text{CH}_3\text{OH} + 3\text{O}_2 \quad (\Delta H_3' = +1430 \text{kJ})
\][/tex]
- Next, take half of the second equation:
[tex]\[
\text{CO} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}_2 \quad \left(\frac{\Delta H_2}{2} = -283 \text{kJ}\right)
\][/tex]
3. Combine the above steps:
- Use the first equation as it is:
[tex]\[
2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \quad (\Delta H_1 = -571 \text{kJ})
\][/tex]
4. Find the overall enthalpy change. According to Hess's Law, sum up the enthalpies:
- For the reverse of equation 3:
[tex]\[
\Delta H_3' = +1430 \text{kJ}
\][/tex]
- For half of equation 2:
[tex]\[
\frac{\Delta H_2}{2} = -283 \text{kJ}
\][/tex]
- For half of equation 1:
[tex]\[
\frac{\Delta H_1}{2} = \frac{-571}{2} = -285.5 \text{kJ}
\][/tex]
5. Sum the enthalpies:
- The total enthalpy change for the reaction:
[tex]\[
\Delta H_{\text{total}} = \left(\frac{\Delta H_1}{2}\right) + \left(\frac{\Delta H_2}{2}\right) + \left(\frac{\Delta H_3}{2}\right)
\][/tex]
- Substituting the values:
[tex]\[
\Delta H_{\text{total}} = -285.5 \text{kJ} + (-283 \text{kJ}) + 715 \text{kJ} = 146.5 \text{kJ}
\][/tex]
Therefore, the enthalpy change for the reaction [tex]\( \text{CO} + 2\text{H}_2 \rightarrow \text{CH}_3\text{OH} \)[/tex] is [tex]\( \boxed{+146.5 \text{kJ}} \)[/tex].
