NO gas is converted to NO₂ gas according to the following reaction:
NO(g) + $\frac { 1 } { 2 }$ O₂(g) $\rightarrow$ NO₂(g)
What is the standard entropy change when 0.5 mol of NO gas reacts with 0.5 mol of O₂ gas? $\begin{array} { l c } \text { Substance } & \boldsymbol { S } ^ { \circ } ( \mathbf { J } / \mathbf { m o l } \cdot \mathbf { K } ) \\\mathrm { NO } ( g ) & 210.7 \\\mathrm { O } _ { 2 } ( g ) & 205.0 \\\mathrm { NO } _ { 2 } ( g ) & 240.0\end{array}$

Question

NO gas is converted to NO₂ gas according to the following reaction:
NO(g) +

\frac { 1 } { 2 }

O₂(g)

\rightarrow

NO₂(g)
What is the standard entropy change when 0.5 mol of NO gas reacts with 0.5 mol of O₂ gas?

\begin{array} { l c } \text { Substance } & \boldsymbol { S } ^ { \circ } ( \mathbf { J } / \mathbf { m o l } \cdot \mathbf { K } ) \\\mathrm { NO } ( g ) & 210.7 \\\mathrm { O } _ { 2 } ( g ) & 205.0 \\\mathrm { NO } _ { 2 } ( g ) & 240.0\end{array}

Quizplus · Accepted Answer

The standard entropy change ($\Delta S^\circ$) for the reaction is calculated using the formula: $\Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}$. For the reaction: $\Delta S^\circ = [240.0] - [210.7 + 0.5 \times 205.0] = -36.6 \, \text{J/K}$.

NO Gas Is Converted to NO2 Gas According to the Following

NO Gas Is Converted to NO₂ Gas According to the Following