madey21
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Calculate the enthalpy for this reaction:

[tex]\[
CH_3Cl + O_2 \rightarrow CO + HCl + H_2O
\][/tex]

Given the following thermochemical equations:

1. [tex]\[2 H_2 + O_2 \rightarrow 2 H_2O, \Delta H_1 = -571 \text{ kJ}\][/tex]
2. [tex]\[CO + 2 H_2 \rightarrow CH_3OH, \Delta H_2 = -139 \text{ kJ}\][/tex]
3. [tex]\[CH_3OH + HCl \rightarrow CH_3Cl + H_2O, \Delta H_3 = -28 \text{ kJ}\][/tex]

[tex]\[
\Delta H_{\text{reaction}} = [?] \text{ kJ}
\][/tex]

Enter either a '+' or '-' sign and the magnitude. Use significant figures.

Sagot :

GinnyAnswer
To calculate the enthalpy change ([tex]\(\Delta H_{rxn}\)[/tex]) for the given reaction:

[tex]\[ \text{CH}_3\text{Cl} + O_2 \rightarrow CO + \text{HCl} + \text{H}_2\text{O} \][/tex]

we will use Hess's Law, which states that the total enthalpy change for a given reaction is the sum of the enthalpy changes of the intermediate steps that lead to that overall reaction.

We will use the following given thermochemical equations and their enthalpy changes ([tex]\(\Delta H\)[/tex]):

1. [tex]\(2 \text{H}_2 + O_2 \rightarrow 2 \text{H}_2\text{O}, \quad \Delta H_1 = -571 \text{ kJ}\)[/tex]
2. [tex]\(\text{CO} + 2 \text{H}_2 \rightarrow \text{CH}_3\text{OH}, \quad \Delta H_2 = -139 \text{ kJ}\)[/tex]
3. [tex]\(\text{CH}_3\text{OH} + \text{HCl} \rightarrow \text{CH}_3\text{Cl} + \text{H}_2\text{O}, \quad \Delta H_3 = -28 \text{ kJ}\)[/tex]

First, we will reverse reaction 3, since we need [tex]\(\text{CH}_3\text{Cl}\)[/tex] as a reactant and not as a product. Reversing the reaction also reverses the sign of [tex]\(\Delta H_3\)[/tex].

[tex]\[ \text{CH}_3\text{Cl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \text{HCl}, \quad \Delta H = +28 \text{ kJ} \][/tex]

Now, we add this reversed [tex]\( \Delta H \)[/tex] value to the [tex]\(\Delta H\)[/tex] values of reactions 1 and 2:

[tex]\[ \begin{align*} 1. & \quad 2 \text{H}_2 + O_2 \rightarrow 2 \text{H}_2\text{O}, \quad \Delta H_1 = -571 \text{ kJ} \\ 2. & \quad \text{CO} + 2 \text{H}_2 \rightarrow \text{CH}_3\text{OH}, \quad \Delta H_2 = -139 \text{ kJ} \\ 3. & \quad \text{CH}_3\text{Cl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \text{HCl}, \quad \Delta H = +28 \text{ kJ} \\ \end{align*} \][/tex]

Summing these up:

[tex]\[ \Delta H_{\text{reaction}} = \Delta H_1 + \Delta H_2 + \Delta H\text{ (reversed 3)} \][/tex]

[tex]\[ \Delta H_{\text{reaction}} = -571 \text{ kJ} + (-139 \text{ kJ}) + (28 \text{ kJ}) \][/tex]

[tex]\[ \Delta H_{\text{reaction}} = -571 - 139 + 28 \text{ kJ} \][/tex]

[tex]\[ \Delta H_{\text{reaction}} = -682 \text{ kJ} \][/tex]

Therefore, the enthalpy change ([tex]\(\Delta H_{\text{r\times n}}\)[/tex]) for the given reaction is:

[tex]\[ \boxed{-682 \text{ kJ}} \][/tex]
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