Question

each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 74.0 °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{co_{2}(g) + 2h_{2}o(g) -> ch_{4}(g) + 2o_{2}(g)}\\) \\(\delta h = 803.\text{ kj}\\) \\(\delta s = 2313.\\ \frac{\text{j}}{\text{k}}\\) \\(\delta g = \square \text{ kj}\\) which is spontaneous? \\(\circ\\) this reaction \\(\circ\\) the reverse reaction \\(\circ\\) neither \\(\ce{2h_{2}(g) + o_{2}(g) -> 2h_{2}o(g)}\\) \\(\delta h = -484.\text{ kj}\\) \\(\delta s = \square \\ \frac{\text{j}}{\text{k}}\\) \\(\delta g = 15.\text{ kj}\\) which is spontaneous? \\(\circ\\) this reaction \\(\circ\\) the reverse reaction

Explanation:

First Reaction: $\boldsymbol{\ce{CO2(g) + 2H2O(g) -> CH4(g) + 2O2(g)}}$

Step 1: Convert temperature to Kelvin

The temperature is $74.0^\circ \text{C}$. To convert to Kelvin, we use the formula $T = T^\circ\text{C} + 273.15$.
So, $T = 74.0 + 273.15 = 347.15\ \text{K}$.

Step 2: Recall the Gibbs free energy formula

The formula for Gibbs free energy is $\Delta G = \Delta H - T\Delta S$. We need to ensure the units of $\Delta H$ and $T\Delta S$ are consistent. $\Delta H$ is in $\text{kJ}$, and $\Delta S$ is in $\frac{\text{J}}{\text{K}}$, so we convert $\Delta S$ to $\frac{\text{kJ}}{\text{K}}$ by dividing by 1000.

$\Delta S = 2313\ \frac{\text{J}}{\text{K}} = 2.313\ \frac{\text{kJ}}{\text{K}}$ (since $1\ \text{kJ} = 1000\ \text{J}$).

Step 3: Calculate $\Delta G$

Substitute $\Delta H = 803\ \text{kJ}$, $T = 347.15\ \text{K}$, and $\Delta S = 2.313\ \frac{\text{kJ}}{\text{K}}$ into the formula:

$$\begin{align*} \Delta G &= \Delta H - T\Delta S \\ &= 803\ \text{kJ} - (347.15\ \text{K} \times 2.313\ \frac{\text{kJ}}{\text{K}}) \\ &= 803\ \text{kJ} - (347.15 \times 2.313)\ \text{kJ} \\ &\approx 803\ \text{kJ} - 793\ \text{kJ} \\ &= 10\ \text{kJ} \end{align*}$$

Step 4: Determine spontaneity

For a reaction, if $\Delta G > 0$, the reverse reaction is spontaneous (the forward reaction is non - spontaneous). Since $\Delta G = 10\ \text{kJ} > 0$, the reverse reaction is spontaneous.

Second Reaction: $\boldsymbol{\ce{2H2(g) + O2(g) -> 2H2O(g)}}$

Step 1: Recall the Gibbs free energy formula

Again, use $\Delta G = \Delta H - T\Delta S$. We can rearrange this formula to solve for $\Delta S$: $\Delta S = \frac{\Delta H - \Delta G}{T}$.

Step 2: Substitute the known values

We know $\Delta H = - 484\ \text{kJ}$, $\Delta G = 15\ \text{kJ}$, and $T = 347.15\ \text{K}$ (same temperature as the first reaction).
First, calculate $\Delta H - \Delta G$:
$\Delta H - \Delta G=-484\ \text{kJ}-15\ \text{kJ}=-499\ \text{kJ}$

Then, solve for $\Delta S$:

$$\begin{align*} \Delta S&=\frac{\Delta H - \Delta G}{T}\\ &=\frac{- 499\ \text{kJ}}{347.15\ \text{K}}\\ &\approx - 1.44\ \frac{\text{kJ}}{\text{K}} \end{align*}$$

Convert $\Delta S$ to $\frac{\text{J}}{\text{K}}$ by multiplying by 1000:
$\Delta S\approx - 1.44\times1000\ \frac{\text{J}}{\text{K}}=-1440\ \frac{\text{J}}{\text{K}}$ (rounded to zero decimal places).

Step 3: Determine spontaneity

For the reaction $\ce{2H2(g) + O2(g) -> 2H2O(g)}$, $\Delta G = 15\ \text{kJ}>0$, so the forward reaction is non - spontaneous. This means the reverse reaction is spontaneous.

Final Answers:

• For the first reaction:

$\Delta G = \boldsymbol{10}\ \text{kJ}$, and the spontaneous reaction is the reverse reaction.

• For the second reaction:

$\Delta S=\boldsymbol{-1440}\ \frac{\text{J}}{\text{K}}$, and the spontaneous reaction is the reverse reaction.

Answer:

First Reaction: $\boldsymbol{\ce{CO2(g) + 2H2O(g) -> CH4(g) + 2O2(g)}}$

Step 1: Convert temperature to Kelvin

The temperature is $74.0^\circ \text{C}$. To convert to Kelvin, we use the formula $T = T^\circ\text{C} + 273.15$.
So, $T = 74.0 + 273.15 = 347.15\ \text{K}$.

Step 2: Recall the Gibbs free energy formula

The formula for Gibbs free energy is $\Delta G = \Delta H - T\Delta S$. We need to ensure the units of $\Delta H$ and $T\Delta S$ are consistent. $\Delta H$ is in $\text{kJ}$, and $\Delta S$ is in $\frac{\text{J}}{\text{K}}$, so we convert $\Delta S$ to $\frac{\text{kJ}}{\text{K}}$ by dividing by 1000.

$\Delta S = 2313\ \frac{\text{J}}{\text{K}} = 2.313\ \frac{\text{kJ}}{\text{K}}$ (since $1\ \text{kJ} = 1000\ \text{J}$).

Step 3: Calculate $\Delta G$

Substitute $\Delta H = 803\ \text{kJ}$, $T = 347.15\ \text{K}$, and $\Delta S = 2.313\ \frac{\text{kJ}}{\text{K}}$ into the formula:

$$\begin{align*} \Delta G &= \Delta H - T\Delta S \\ &= 803\ \text{kJ} - (347.15\ \text{K} \times 2.313\ \frac{\text{kJ}}{\text{K}}) \\ &= 803\ \text{kJ} - (347.15 \times 2.313)\ \text{kJ} \\ &\approx 803\ \text{kJ} - 793\ \text{kJ} \\ &= 10\ \text{kJ} \end{align*}$$

Step 4: Determine spontaneity

For a reaction, if $\Delta G > 0$, the reverse reaction is spontaneous (the forward reaction is non - spontaneous). Since $\Delta G = 10\ \text{kJ} > 0$, the reverse reaction is spontaneous.

Second Reaction: $\boldsymbol{\ce{2H2(g) + O2(g) -> 2H2O(g)}}$

Step 1: Recall the Gibbs free energy formula

Again, use $\Delta G = \Delta H - T\Delta S$. We can rearrange this formula to solve for $\Delta S$: $\Delta S = \frac{\Delta H - \Delta G}{T}$.

Step 2: Substitute the known values

We know $\Delta H = - 484\ \text{kJ}$, $\Delta G = 15\ \text{kJ}$, and $T = 347.15\ \text{K}$ (same temperature as the first reaction).
First, calculate $\Delta H - \Delta G$:
$\Delta H - \Delta G=-484\ \text{kJ}-15\ \text{kJ}=-499\ \text{kJ}$

Then, solve for $\Delta S$:

$$\begin{align*} \Delta S&=\frac{\Delta H - \Delta G}{T}\\ &=\frac{- 499\ \text{kJ}}{347.15\ \text{K}}\\ &\approx - 1.44\ \frac{\text{kJ}}{\text{K}} \end{align*}$$

Convert $\Delta S$ to $\frac{\text{J}}{\text{K}}$ by multiplying by 1000:
$\Delta S\approx - 1.44\times1000\ \frac{\text{J}}{\text{K}}=-1440\ \frac{\text{J}}{\text{K}}$ (rounded to zero decimal places).

Step 3: Determine spontaneity

For the reaction $\ce{2H2(g) + O2(g) -> 2H2O(g)}$, $\Delta G = 15\ \text{kJ}>0$, so the forward reaction is non - spontaneous. This means the reverse reaction is spontaneous.

Final Answers:

• For the first reaction:

$\Delta G = \boldsymbol{10}\ \text{kJ}$, and the spontaneous reaction is the reverse reaction.

• For the second reaction:

$\Delta S=\boldsymbol{-1440}\ \frac{\text{J}}{\text{K}}$, and the spontaneous reaction is the reverse reaction.