Questions: Calculate the standard-state entropy for the following reaction: 1 CH4(g) + 2 O2(g) -> 1 CO2(g) + 2 H2O(l) (If applicable, coefficients of one have been included for clarity.) The standard entropy values are given in the table. Formula S° J /(K · mol) CO2(g) 214 H2O(l) 189 CH4(g) 186 O2(g) 205 Express your answer with the appropriate units.

Calculate the standard-state entropy for the following reaction:
1 CH4(g) + 2 O2(g) -> 1 CO2(g) + 2 H2O(l)
(If applicable, coefficients of one have been included for clarity.)
The standard entropy values are given in the table.
Formula S° J /(K · mol)
CO2(g) 214
H2O(l) 189
CH4(g) 186
O2(g) 205

Express your answer with the appropriate units.

Transcript text: Calculate the standard-state entropy for the following reaction: \[ 1 \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 1 \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \] (If applicable, coefficients of one have been included for clarity.) The standard entropy values are given in the table. \begin{tabular}{|c|c|} \hline Formula & \begin{tabular}{c} $S^{\circ}$ \\ $\mathrm{J} /(\mathrm{K} \cdot \mathrm{mol})$ \end{tabular} \\ \hline $\mathrm{CO}_{2}(\mathrm{~g})$ & 214 \\ \hline $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ & 189 \\ \hline $\mathrm{CH}_{4}(\mathrm{~g})$ & 186 \\ \hline $\mathrm{O}_{2}(\mathrm{~g})$ & 205 \\ \hline \end{tabular} Express your answer with the appropriate units. View Available Hint(s) $\square$ $\Delta S^{\circ}=$ $\square$ Value Units Submit

Solution

Solution Steps

Step 1: Understand the Reaction and Given Data

The reaction given is:

\[ 1 \mathrm{CH}_{4}(\mathrm{~g}) + 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 1 \mathrm{CO}_{2}(\mathrm{~g}) + 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \]

We are provided with the standard entropy values ($S^{\circ}$) for each substance involved in the reaction:

$S^{\circ}(\mathrm{CO}_{2}(\mathrm{~g})) = 214 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol})$
$S^{\circ}(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})) = 189 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol})$
$S^{\circ}(\mathrm{CH}_{4}(\mathrm{~g})) = 186 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol})$
$S^{\circ}(\mathrm{O}_{2}(\mathrm{~g})) = 205 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol})$

Step 2: Calculate the Standard-State Entropy Change

The standard-state entropy change ($\Delta S^{\circ}$) for the reaction is calculated using the formula:

\[ \Delta S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants}) \]

Step 3: Calculate the Entropy of Products

For the products, we have:

1 mole of $\mathrm{CO}_{2}(\mathrm{~g})$: $1 \times 214 = 214 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol})$
2 moles of $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$: $2 \times 189 = 378 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol})$

Total entropy of products:

\[ 214 + 378 = 592 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol}) \]

Step 4: Calculate the Entropy of Reactants

For the reactants, we have:

1 mole of $\mathrm{CH}_{4}(\mathrm{~g})$: $1 \times 186 = 186 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol})$
2 moles of $\mathrm{O}_{2}(\mathrm{~g})$: $2 \times 205 = 410 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol})$

Total entropy of reactants:

\[ 186 + 410 = 596 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol}) \]

Step 5: Calculate $\Delta S^{\circ}$

Now, calculate the standard-state entropy change:

\[ \Delta S^{\circ} = 592 - 596 = -4 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol}) \]