Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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 :

GinnyAnswer
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.
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.