Madey21idn
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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]\[
\Delta H_{\text{rxn}} = [?] \, \text{kJ}
\][/tex]

Enter either a [tex]\(+\)[/tex] or [tex]\(-\)[/tex] sign and the magnitude. Use significant figures.

Sagot :

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

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

we need to manipulate the given thermochemical equations:

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

First, we reverse equation 3 to match our desired reactants and products:

[tex]\[ \text{CH}_3\text{Cl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \text{HCl}, \Delta H_{\text{reverse 3}} = +28 \, \text{kJ} \][/tex]

Next, we combine equations to achieve the target reaction. We add equation 1 and the reversed equation 3:

[tex]\[ \begin{align*} & \text{CH}_3\text{Cl} + \text{H}_2\text{O} + 2 \text{H}_2 + \text{O}_2 \rightarrow \text{CH}_3\text{OH} + \text{HCl} + 2 \text{H}_2\text{O} \\ & \Delta H = -571 \, \text{kJ} + 28 \, \text{kJ} = -543 \, \text{kJ} \end{align*} \][/tex]

Then, add equation 2 to this resulting equation:

[tex]\[ \begin{align*} & \text{CH}_3\text{Cl} + \text{H}_2\text{O} + 2 \text{H}_2 + \text{O}_2 + \text{CO} + 2\text{H}_2 \rightarrow \text{CH}_3\text{OH} + \text{HCl} + 2\text{H}_2\text{O} + \text{CH}_3\text{OH} \\ & \Delta H = -543 \, \text{kJ} - 139 \, \text{kJ} = -682 \, \text{kJ} \end{align*} \][/tex]

Thus, the enthalpy change for the reaction:

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

is [tex]\(\Delta H_{rxn} = -682 \, \text{kJ}\)[/tex]
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