Answered

Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

The decomposition of calcium carbonate, [tex]\[ CaCO_3(s) \rightarrow CaO(s) + CO_2(g) \][/tex], has the following values for free energy and enthalpy at [tex]\[ 25.0^{\circ}C \][/tex].

[tex]\[
\begin{array}{l}
\Delta G = 130.5 \, \text{kJ/mol} \\
\Delta H = 178.3 \, \text{kJ/mol}
\end{array}
\][/tex]

What is the entropy of the reaction? Use [tex]\[ \Delta G = \Delta H - T \Delta S \][/tex].

A. [tex]\[-160.3 \, \text{J/(mol·K)}\][/tex]
B. [tex]\[-47.8 \, \text{J/(mol·K)}\][/tex]
C. [tex]\[160.3 \, \text{J/(mol·K)}\][/tex]

Sagot :

GinnyAnswer
To solve for the entropy change [tex]\(\Delta S\)[/tex] of the reaction, we will use the given Gibbs free energy change [tex]\(\Delta G\)[/tex], enthalpy change [tex]\(\Delta H\)[/tex], and the temperature [tex]\(T\)[/tex] along with the formula [tex]\(\Delta G = \Delta H - T\Delta S\)[/tex],

Here are the given values:
[tex]\[ \Delta G = 130.5 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H = 178.3 \, \text{kJ/mol} \][/tex]
[tex]\[ T = 25.0^\circ \text{C} = 25.0 + 273.15 \, \text{K} = 298.15 \, \text{K} \][/tex]

We need to rearrange the formula [tex]\(\Delta G = \Delta H - T\Delta S\)[/tex] to solve for [tex]\(\Delta S\)[/tex]:
[tex]\[ \Delta S = \frac{\Delta H - \Delta G}{T} \][/tex]

First, substitute the given values into the equation. Since [tex]\(\Delta H\)[/tex] and [tex]\(\Delta G\)[/tex] are given in kJ/mol, we need to convert the result into J/(mol·K) by multiplying by 1000.

[tex]\[ \Delta S = \frac{(178.3 - 130.5) \, \text{kJ/mol}}{298.15 \, \text{K}} \cdot 1000 \, \frac{\text{J}}{\text{kJ}} \][/tex]

Calculate the difference in enthalpy and free energy:
[tex]\[ 178.3 \, \text{kJ/mol} - 130.5 \, \text{kJ/mol} = 47.8 \, \text{kJ/mol} \][/tex]

Now, substitute this value into the equation:
[tex]\[ \Delta S = \frac{47.8 \, \text{kJ/mol}}{298.15 \, \text{K}} \cdot 1000 \, \frac{\text{J}}{\text{kJ}} \][/tex]

[tex]\[ \Delta S = \frac{47800 \, \text{J/mol}}{298.15 \, \text{K}} \][/tex]

Finally, we perform the division to find [tex]\(\Delta S\)[/tex]:
[tex]\[ \Delta S \approx 160.32 \, \text{J/(mol·K)} \][/tex]

Therefore, the entropy change [tex]\(\Delta S\)[/tex] of the reaction is approximately [tex]\(160.3 \, \text{J/(mol·K)}\)[/tex].

Hence, the correct answer is:

[tex]\[\boxed{160.3 \, \text{J/(mol·K)}}\][/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.