Question

each case, she fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 91.0 °c and constant total pressure. then, she 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 her 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. \\(2\text{al}(s) + 3\text{h}_2\text{o}(g) \
ightarrow \text{al}_2\text{o}_3(s) + 3\text{h}_2(g)\\) \\(\delta h = -951\text{ kj}\\) \\(\delta s = -2612 \frac{\text{j}}{\text{k}}\\) \\(\delta g = \square \text{ kj}\\) which is spontaneous? \\(\circ\\) this reaction \\(\circ\\) the reverse reaction \\(\circ\\) neither \\(\text{n}_2(g) + \text{o}_2(g) \
ightarrow 2\text{no}(g)\\) \\(\delta h = 181\text{ kj}\\) \\(\delta s = \square \frac{\text{j}}{\text{k}}\\) \\(\delta g = -22\text{ kj}\\) which is spontaneous? \\(\circ\\) this reaction \\(\circ\\) the reverse reaction

Explanation:

First Reaction: \( \boldsymbol{2\text{Al}(s) + 3\text{H}_2\text{O}(g)

ightarrow \text{Al}_2\text{O}_3(s) + 3\text{H}_2(g)} \)

Step 1: Convert temperature to Kelvin

The temperature is \( 91.0^\circ \text{C} \). To convert to Kelvin, use \( T = 91.0 + 273.15 = 364.15 \, \text{K} \).

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

Given \( \Delta H = -951 \, \text{kJ} = -951000 \, \text{J} \), \( \Delta S = -2612 \, \frac{\text{J}}{\text{K}} \), and \( T = 364.15 \, \text{K} \).

First, calculate \( T\Delta S \):
\( T\Delta S = 364.15 \, \text{K} \times (-2612 \, \frac{\text{J}}{\text{K}}) = -364.15 \times 2612 \, \text{J} \approx -951159.8 \, \text{J} \approx -951.16 \, \text{kJ} \)

Now, calculate \( \Delta G \):
\( \Delta G = \Delta H - T\Delta S = -951000 \, \text{J} - (-951159.8 \, \text{J}) = -951000 + 951159.8 = 159.8 \, \text{J} \approx 0 \, \text{kJ} \) (rounded to zero decimal places). Wait, let's check the calculation again. Wait, \( \Delta H \) is in kJ, \( \Delta S \) is in J/K. Let's do unit conversion properly.

\( \Delta H = -951 \, \text{kJ} = -951000 \, \text{J} \)

\( T\Delta S = 364.15 \, \text{K} \times (-2612 \, \text{J/K}) = -364.15 \times 2612 = -951159.8 \, \text{J} = -951.16 \, \text{kJ} \)

Then \( \Delta G = \Delta H - T\Delta S = -951 \, \text{kJ} - (-951.16 \, \text{kJ}) = -951 + 951.16 = 0.16 \, \text{kJ} \approx 0 \, \text{kJ} \) (rounded to zero decimal places). Wait, maybe I made a mistake. Wait, \( \Delta H \) is -951 kJ, \( \Delta S \) is -2612 J/K. Let's compute \( T\Delta S \) in kJ:

\( T = 91 + 273.15 = 364.15 \, \text{K} \)

\( T\Delta S = 364.15 \, \text{K} \times (-2612 \, \text{J/K}) = -364.15 \times 2612 \, \text{J} = -951159.8 \, \text{J} = -951.16 \, \text{kJ} \)

Then \( \Delta G = \Delta H - T\Delta S = -951 \, \text{kJ} - (-951.16 \, \text{kJ}) = 0.16 \, \text{kJ} \approx 0 \, \text{kJ} \) (rounded to zero decimal places). So \( \Delta G \approx 0 \, \text{kJ} \).

Step 4: Determine spontaneity

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

Second Reaction: \( \boldsymbol{\text{N}_2(g) + \text{O}_2(g)

ightarrow 2\text{NO}(g)} \)

Step 1: Recall the Gibbs free energy formula

Again, use \( \Delta G = \Delta H - T\Delta S \). We need to solve for \( \Delta S \).

Step 2: Rearrange the formula

\( \Delta S = \frac{\Delta H - \Delta G}{T} \)

Step 3: Substitute the values

Given \( \Delta H = 181 \, \text{kJ} = 181000 \, \text{J} \), \( \Delta G = -22 \, \text{kJ} = -22000 \, \text{J} \), and \( T = 364.15 \, \text{K} \).

First, calculate \( \Delta H - \Delta G \):
\( \Delta H - \Delta G = 181000 \, \text{J} - (-22000 \, \text{J}) = 181000 + 22000 = 203000 \, \text{J} \)

Then, calculate \( \Delta S \):
\( \Delta S = \frac{203000 \, \text{J}}{364.15 \, \text{K}} \approx 557.46 \, \frac{\text{J}}{\text{K}} \approx 557 \, \frac{\text{J}}{\text{K}} \) (rounded to zero decimal places).

Step 4: Determine spontaneity

Given \( \Delta G = -22 \, \text{kJ} < 0 \), the forward reaction (this reaction) is spontaneous.

Final Answers

First Reaction:

• \( \Delta G = \boxed{0} \, \text{kJ} \)
• Spontaneity: neither

Second Reaction:

• \( \Delta S = \boxed{557} \, \frac{\text{J}}{\text{K}} \)
• Spontaneity: this reaction

Answer:

First Reaction: \( \boldsymbol{2\text{Al}(s) + 3\text{H}_2\text{O}(g)

ightarrow \text{Al}_2\text{O}_3(s) + 3\text{H}_2(g)} \)

Step 1: Convert temperature to Kelvin

The temperature is \( 91.0^\circ \text{C} \). To convert to Kelvin, use \( T = 91.0 + 273.15 = 364.15 \, \text{K} \).

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

Given \( \Delta H = -951 \, \text{kJ} = -951000 \, \text{J} \), \( \Delta S = -2612 \, \frac{\text{J}}{\text{K}} \), and \( T = 364.15 \, \text{K} \).

First, calculate \( T\Delta S \):
\( T\Delta S = 364.15 \, \text{K} \times (-2612 \, \frac{\text{J}}{\text{K}}) = -364.15 \times 2612 \, \text{J} \approx -951159.8 \, \text{J} \approx -951.16 \, \text{kJ} \)

Now, calculate \( \Delta G \):
\( \Delta G = \Delta H - T\Delta S = -951000 \, \text{J} - (-951159.8 \, \text{J}) = -951000 + 951159.8 = 159.8 \, \text{J} \approx 0 \, \text{kJ} \) (rounded to zero decimal places). Wait, let's check the calculation again. Wait, \( \Delta H \) is in kJ, \( \Delta S \) is in J/K. Let's do unit conversion properly.

\( \Delta H = -951 \, \text{kJ} = -951000 \, \text{J} \)

\( T\Delta S = 364.15 \, \text{K} \times (-2612 \, \text{J/K}) = -364.15 \times 2612 = -951159.8 \, \text{J} = -951.16 \, \text{kJ} \)

Then \( \Delta G = \Delta H - T\Delta S = -951 \, \text{kJ} - (-951.16 \, \text{kJ}) = -951 + 951.16 = 0.16 \, \text{kJ} \approx 0 \, \text{kJ} \) (rounded to zero decimal places). Wait, maybe I made a mistake. Wait, \( \Delta H \) is -951 kJ, \( \Delta S \) is -2612 J/K. Let's compute \( T\Delta S \) in kJ:

\( T = 91 + 273.15 = 364.15 \, \text{K} \)

\( T\Delta S = 364.15 \, \text{K} \times (-2612 \, \text{J/K}) = -364.15 \times 2612 \, \text{J} = -951159.8 \, \text{J} = -951.16 \, \text{kJ} \)

Then \( \Delta G = \Delta H - T\Delta S = -951 \, \text{kJ} - (-951.16 \, \text{kJ}) = 0.16 \, \text{kJ} \approx 0 \, \text{kJ} \) (rounded to zero decimal places). So \( \Delta G \approx 0 \, \text{kJ} \).

Step 4: Determine spontaneity

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

Second Reaction: \( \boldsymbol{\text{N}_2(g) + \text{O}_2(g)

ightarrow 2\text{NO}(g)} \)

Step 1: Recall the Gibbs free energy formula

Again, use \( \Delta G = \Delta H - T\Delta S \). We need to solve for \( \Delta S \).

Step 2: Rearrange the formula

\( \Delta S = \frac{\Delta H - \Delta G}{T} \)

Step 3: Substitute the values

Given \( \Delta H = 181 \, \text{kJ} = 181000 \, \text{J} \), \( \Delta G = -22 \, \text{kJ} = -22000 \, \text{J} \), and \( T = 364.15 \, \text{K} \).

First, calculate \( \Delta H - \Delta G \):
\( \Delta H - \Delta G = 181000 \, \text{J} - (-22000 \, \text{J}) = 181000 + 22000 = 203000 \, \text{J} \)

Then, calculate \( \Delta S \):
\( \Delta S = \frac{203000 \, \text{J}}{364.15 \, \text{K}} \approx 557.46 \, \frac{\text{J}}{\text{K}} \approx 557 \, \frac{\text{J}}{\text{K}} \) (rounded to zero decimal places).

Step 4: Determine spontaneity

Given \( \Delta G = -22 \, \text{kJ} < 0 \), the forward reaction (this reaction) is spontaneous.

Final Answers

First Reaction:

• \( \Delta G = \boxed{0} \, \text{kJ} \)
• Spontaneity: neither

Second Reaction:

• \( \Delta S = \boxed{557} \, \frac{\text{J}}{\text{K}} \)
• Spontaneity: this reaction