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Consider the intermediate equations:
[tex]\[
\begin{array}{ll}
C(s) + O_2(g) \rightarrow CO_2(g) & \Delta H_1 = -393.5 \, \text{kJ} \\
2 \, CO(g) + O_2(g) \rightarrow 2 \, CO_2(g) & \Delta H_2 = -566.0 \, \text{kJ} \\
2 \, H_2O(g) \rightarrow 2 \, H_2(g) + O_2(g) & \Delta H_3 = 483.6 \, \text{kJ}
\end{array}
\][/tex]

With the overall reaction:
[tex]\[
C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \quad \Delta H_{mn} = \text{ ? }
\][/tex]

What must be done to calculate the enthalpy of reaction? Check all that apply.

- The first equation must be halved.
- The first equation must be reversed.
- The second equation must be halved.
- The second equation must be reversed.
- The third equation must be halved.
- The third equation must be reversed.

What is the overall enthalpy of reaction?
[tex]\[
\Delta H_{xn} = \square \, \text{kJ}
\][/tex]

Sagot :

GinnyAnsweridn
To calculate the enthalpy of the overall reaction
[tex]\[ C(s) + H_2O(g) \rightarrow CO(g) + H_2(g), \][/tex]
we need to manipulate the given intermediate equations in such a way that they sum up to the overall reaction.

Intermediate Equations and Steps:

1. First Equation:
[tex]\[ C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta H_1 = -393.5 \text{ kJ} \][/tex]
To use this equation in the overall reaction, we need to reverse it to produce [tex]\( C(s) \)[/tex] and [tex]\( O_2(g) \)[/tex] from [tex]\( CO_2(g) \)[/tex]:
[tex]\[ CO_2(g) \rightarrow C(s) + O_2(g) \][/tex]
Reversing the equation changes the sign of [tex]\(\Delta H_1\)[/tex]:
[tex]\[ \Delta H_1 = +393.5 \text{ kJ} \][/tex]

2. Second Equation:
[tex]\[ 2 CO(g) + O_2(g) \rightarrow 2 CO_2(g), \quad \Delta H_2 = -566.0 \text{ kJ} \][/tex]
We need to produce [tex]\( CO(g) \)[/tex] on the product side, so let's halve the equation first:
[tex]\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \][/tex]
Halving the equation also halves the enthalpy change:
[tex]\[ \Delta H_2 = \frac{-566.0}{2} = -283.0 \text{ kJ} \][/tex]

3. Third Equation:
[tex]\[ 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g), \quad \Delta H_3 = 483.6 \text{ kJ} \][/tex]
To balance the overall reaction, we also need to halve this equation:
[tex]\[ H_2O(g) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \][/tex]
Halving this equation halves the enthalpy change:
[tex]\[ \Delta H_3 = \frac{483.6}{2} = 241.8 \text{ kJ} \][/tex]

Next, Sum up the modified enthalpies to get the overall reaction enthalpy.

Modified enthalpies:
[tex]\[ \Delta H_1 = +393.5 \text{ kJ} \][/tex]
[tex]\[ \Delta H_2 = -283.0 \text{ kJ} \][/tex]
[tex]\[ \Delta H_3 = 241.8 \text{ kJ} \][/tex]

Now, summing these values:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 + \Delta H_3 = 393.5 \text{ kJ} - 283.0 \text{ kJ} + 241.8 \text{ kJ} = 352.3 \text{ kJ} \][/tex]

Therefore, the enthalpy change for the overall reaction is:
[tex]\[ \Delta H_{\text{xn}} = 352.3 \text{ kJ} \][/tex]

Necessary Steps to Calculate:
- The first equation must be reversed.
- The second equation must be halved.
- The third equation must be halved.

These are the steps required to determine the overall enthalpy change.
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