Question

each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 35.0 °c and constant total pressure. then, he measures the reaction enthalpy δh and reaction entropy δs of the first reaction, and the reaction enthalpy δh and reaction free energy δg of the second reaction. the results of his measurements are shown in the table. complete the table. that is, calculate δg for the first reaction and δs 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. \\(\ce{c6h12o6(s) -> 6c(s) + 6h2(g) + 3o2(g)}\\) \\(\delta h = 1237\\,\text{kj}\\) \\(\delta s = 3922\\,\frac{\text{j}}{\text{k}}\\) \\(\delta g = \square\\,\text{kj}\\) which is spontaneous? \\(\circ\\) this reaction \\(\circ\\) the reverse reaction \\(\circ\\) neither \\(\ce{hch3co2(l) -> ch3oh(g) + co(g)}\\) \\(\delta h = 172\\,\text{kj}\\) \\(\delta s = \square\\,\frac{\text{j}}{\text{k}}\\) \\(\delta g = 0\\,\text{kj}\\) which is spontaneous? \\(\circ\\) this reaction \\(\circ\\) the reverse reaction

Explanation:

First Reaction: \( \boldsymbol{\mathrm{C_6H_{12}O_6}(s)

ightarrow 6\mathrm{C}(s) + 6\mathrm{H_2}(g) + 3\mathrm{O_2}(g)} \)

Step 1: Convert temperature to Kelvin

The temperature is \( 35.0^\circ \text{C} \). To convert to Kelvin, use \( T = 35.0 + 273.15 = 308.15 \, \text{K} \).

Step 2: Recall the Gibbs free energy formula

The formula for Gibbs free energy is \( \Delta G = \Delta H - T\Delta S \).

Step 3: Substitute the values

We have \( \Delta H = 1237 \, \text{kJ} \), \( \Delta S = 3922 \, \frac{\text{J}}{\text{K}} = 3.922 \, \frac{\text{kJ}}{\text{K}} \) (since \( 1 \, \text{kJ} = 1000 \, \text{J} \)), and \( T = 308.15 \, \text{K} \).

Substitute into the formula:

$$ \Delta G = 1237 \, \text{kJ} - (308.15 \, \text{K} \times 3.922 \, \frac{\text{kJ}}{\text{K}}) $$

Step 4: Calculate \( T\Delta S \)

$$ 308.15 \times 3.922 \approx 308.15 \times 3.922 \approx 1208.5 $$

Step 5: Calculate \( \Delta G \)

$$ \Delta G = 1237 - 1208.5 = 28.5 \, \text{kJ} \approx 29 \, \text{kJ} \, (\text{rounded to zero decimal places}) $$

Step 6: Determine spontaneity

Since \( \Delta G > 0 \), the forward reaction is non - spontaneous, and the reverse reaction is spontaneous.

Second Reaction: \( \boldsymbol{\mathrm{HCH_3CO_2}(l)

ightarrow \mathrm{CH_3OH}(g) + \mathrm{CO}(g)} \)

Step 1: Recall the Gibbs free energy formula

Again, use \( \Delta G = \Delta H - T\Delta S \). We know \( \Delta G = 0 \, \text{kJ} \) (at equilibrium), \( \Delta H = 172 \, \text{kJ} \), and \( T = 308.15 \, \text{K} \).

Step 2: Rearrange the formula to solve for \( \Delta S \)

From \( \Delta G = \Delta H - T\Delta S \), when \( \Delta G = 0 \), we have \( \Delta H = T\Delta S \), so \( \Delta S=\frac{\Delta H}{T} \).

Step 3: Substitute the values

Substitute \( \Delta H = 172 \, \text{kJ}=172000 \, \text{J} \) and \( T = 308.15 \, \text{K} \) into the formula:

$$ \Delta S=\frac{172000 \, \text{J}}{308.15 \, \text{K}}\approx 558.2 \, \frac{\text{J}}{\text{K}} $$

Step 4: Determine spontaneity

Since \( \Delta G = 0 \), the system is at equilibrium, so neither the forward nor the reverse reaction is spontaneous.

Final Answers

• For the first reaction: \( \Delta G=\boldsymbol{29 \, \text{kJ}} \), and the reverse reaction is spontaneous.
• For the second reaction: \( \Delta S=\boldsymbol{558 \, \frac{\text{J}}{\text{K}}} \) (rounded to a reasonable value, or more precisely \( \approx 558 \)), and neither reaction is spontaneous.

Answer:

First Reaction: \( \boldsymbol{\mathrm{C_6H_{12}O_6}(s)

ightarrow 6\mathrm{C}(s) + 6\mathrm{H_2}(g) + 3\mathrm{O_2}(g)} \)

Step 1: Convert temperature to Kelvin

The temperature is \( 35.0^\circ \text{C} \). To convert to Kelvin, use \( T = 35.0 + 273.15 = 308.15 \, \text{K} \).

Step 2: Recall the Gibbs free energy formula

The formula for Gibbs free energy is \( \Delta G = \Delta H - T\Delta S \).

Step 3: Substitute the values

We have \( \Delta H = 1237 \, \text{kJ} \), \( \Delta S = 3922 \, \frac{\text{J}}{\text{K}} = 3.922 \, \frac{\text{kJ}}{\text{K}} \) (since \( 1 \, \text{kJ} = 1000 \, \text{J} \)), and \( T = 308.15 \, \text{K} \).

Substitute into the formula:

$$ \Delta G = 1237 \, \text{kJ} - (308.15 \, \text{K} \times 3.922 \, \frac{\text{kJ}}{\text{K}}) $$

Step 4: Calculate \( T\Delta S \)

$$ 308.15 \times 3.922 \approx 308.15 \times 3.922 \approx 1208.5 $$

Step 5: Calculate \( \Delta G \)

$$ \Delta G = 1237 - 1208.5 = 28.5 \, \text{kJ} \approx 29 \, \text{kJ} \, (\text{rounded to zero decimal places}) $$

Step 6: Determine spontaneity

Since \( \Delta G > 0 \), the forward reaction is non - spontaneous, and the reverse reaction is spontaneous.

Second Reaction: \( \boldsymbol{\mathrm{HCH_3CO_2}(l)

ightarrow \mathrm{CH_3OH}(g) + \mathrm{CO}(g)} \)

Step 1: Recall the Gibbs free energy formula

Again, use \( \Delta G = \Delta H - T\Delta S \). We know \( \Delta G = 0 \, \text{kJ} \) (at equilibrium), \( \Delta H = 172 \, \text{kJ} \), and \( T = 308.15 \, \text{K} \).

Step 2: Rearrange the formula to solve for \( \Delta S \)

From \( \Delta G = \Delta H - T\Delta S \), when \( \Delta G = 0 \), we have \( \Delta H = T\Delta S \), so \( \Delta S=\frac{\Delta H}{T} \).

Step 3: Substitute the values

Substitute \( \Delta H = 172 \, \text{kJ}=172000 \, \text{J} \) and \( T = 308.15 \, \text{K} \) into the formula:

$$ \Delta S=\frac{172000 \, \text{J}}{308.15 \, \text{K}}\approx 558.2 \, \frac{\text{J}}{\text{K}} $$

Step 4: Determine spontaneity

Since \( \Delta G = 0 \), the system is at equilibrium, so neither the forward nor the reverse reaction is spontaneous.

Final Answers

• For the first reaction: \( \Delta G=\boldsymbol{29 \, \text{kJ}} \), and the reverse reaction is spontaneous.
• For the second reaction: \( \Delta S=\boldsymbol{558 \, \frac{\text{J}}{\text{K}}} \) (rounded to a reasonable value, or more precisely \( \approx 558 \)), and neither reaction is spontaneous.