A chemical engineer is studying the two reactions shown in the table below. In each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 20.0 °C and constant total pressure. Then, he measures the reaction enthalpy AH and reaction entropy AS of the first reaction, and the reaction enthalpy AH and reaction free energy AG of the second reaction. The results of his measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. AH-1212. kJ J AS = -4174. K AG = 4H₂O₂ (1) + PbS (s) PbSO4 (s) + 4H₂O (1) | kJ Which is spontaneous? this reaction the reverse reaction neither AH1237. kJ AS = AG = -28. kJ 6C(s) + 6H₂(g) + 30₂ (8) C6H₁2O6 (s) -0-X/2 Which is spontaneous? this reaction the reverse reaction neither 00 3 ?

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### Study of Reaction Thermodynamics

A chemical engineer is studying two reactions, the details of which are outlined in the table below.

#### Experimental Procedure
1. The engineer fills a reaction vessel with a mixture of the reactants and products, maintaining a constant temperature of 20.0 °C and constant total pressure.
2. The reaction enthalpy (ΔH) and entropy (ΔS) for the first reaction are measured.
3. Similarly, the reaction enthalpy (ΔH) and reaction free energy (ΔG) for the second reaction are also measured.

### Table of Measurements
The following table displays the gathered data for the two reactions:

\[
\begin{array}{|l|c|c|}
\hline
\text{Reaction} & 4 \text{H}_2\text{O}_2(\ell) + \text{PbS}(\text{s}) \rightarrow \text{PbSO}_4(\text{s}) + 4 \text{H}_2\text{O}(\ell) & 6 \text{C}(\text{s}) + 6 \text{H}_2(\text{g}) + 3 \text{O}_2(\text{g}) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(\text{s}) \\
\hline
\Delta H & -1212 \, \text{kJ} & -1237 \, \text{kJ} \\
\hline
\Delta S & -4174 \, \frac{\text{J}}{\text{K}} & \boxed{} \\
\hline
\Delta G & \boxed{} & -28 \, \text{kJ} \\
\hline
\text{Which is spontaneous?} & \begin{array}{c}
\text{this reaction} \\
\text{the reverse reaction} \\
\text{neither} \\
\end{array} & \begin{array}{c}
\text{this reaction} \\
\text{the reverse reaction} \\
\text{neither} \\
\end{array} \\
\hline
\end{array}
\]

### Tasks
1. **Calculation**: Determine ΔG for the first reaction and ΔS for the second reaction. Use the Gibbs free energy equation:
\[
Transcribed Image Text:### Study of Reaction Thermodynamics A chemical engineer is studying two reactions, the details of which are outlined in the table below. #### Experimental Procedure 1. The engineer fills a reaction vessel with a mixture of the reactants and products, maintaining a constant temperature of 20.0 °C and constant total pressure. 2. The reaction enthalpy (ΔH) and entropy (ΔS) for the first reaction are measured. 3. Similarly, the reaction enthalpy (ΔH) and reaction free energy (ΔG) for the second reaction are also measured. ### Table of Measurements The following table displays the gathered data for the two reactions: \[ \begin{array}{|l|c|c|} \hline \text{Reaction} & 4 \text{H}_2\text{O}_2(\ell) + \text{PbS}(\text{s}) \rightarrow \text{PbSO}_4(\text{s}) + 4 \text{H}_2\text{O}(\ell) & 6 \text{C}(\text{s}) + 6 \text{H}_2(\text{g}) + 3 \text{O}_2(\text{g}) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(\text{s}) \\ \hline \Delta H & -1212 \, \text{kJ} & -1237 \, \text{kJ} \\ \hline \Delta S & -4174 \, \frac{\text{J}}{\text{K}} & \boxed{} \\ \hline \Delta G & \boxed{} & -28 \, \text{kJ} \\ \hline \text{Which is spontaneous?} & \begin{array}{c} \text{this reaction} \\ \text{the reverse reaction} \\ \text{neither} \\ \end{array} & \begin{array}{c} \text{this reaction} \\ \text{the reverse reaction} \\ \text{neither} \\ \end{array} \\ \hline \end{array} \] ### Tasks 1. **Calculation**: Determine ΔG for the first reaction and ΔS for the second reaction. Use the Gibbs free energy equation: \[
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