Question

each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 54.0 °c and constant total pressure. then, he measures the reaction enthalpy $\delta h$ and reaction entropy $\delta s$ of the first reaction, and the reaction enthalpy $\delta h$ and reaction free energy $\delta g$ of the second reaction. the results of his measurements are shown in the table. complete the table. that is, calculate $\delta g$ for the first reaction and $\delta s$ 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{mgo(s) + 2hcl(g) -> mgcl2(s) + h2o(l)}\\) \\(\delta h = -140\\ \text{kj}\\) \\(\delta s = -330\\ \frac{\text{j}}{\text{k}}\\) \\(\delta g = \square\\ \text{kj}\\) which is spontaneous? \\(\circ\\) this reaction \\(\circ\\) the reverse reaction \\(\circ\\) neither \\(\ce{tio2(s) + 4hcl(g) -> ticl4(g) + 2h2o(g)}\\) \\(\delta h = 70\\ \text{kj}\\) \\(\delta g = 0\\ \text{kj}\\) \\(\delta s = \square\\ \frac{\text{j}}{\text{k}}\\) which is spontaneous? \\(\circ\\) this reaction \\(\circ\\) the reverse reaction \\(\circ\\) neither

Explanation:

First Reaction: $\boldsymbol{\ce{MgO}(s) + 2\ce{HCl}(g)

ightarrow \ce{MgCl_2}(s) + \ce{H_2O}(l)}$

Step 1: Convert temperature to Kelvin

The temperature is $54.0^\circ \text{C}$. To convert to Kelvin, use the formula $T = 273.15 + t$.
$T = 273.15 + 54.0 = 327.15\ \text{K}$

Step 2: Convert $\Delta S$ to kJ/K

$\Delta S = -330\ \frac{\text{J}}{\text{K}} = -0.330\ \frac{\text{kJ}}{\text{K}}$ (since $1\ \text{kJ} = 1000\ \text{J}$)

Step 3: Calculate $\Delta G$ using $\boldsymbol{\Delta G = \Delta H - T\Delta S}$

Given $\Delta H = -140\ \text{kJ}$ and $T = 327.15\ \text{K}$, $\Delta S = -0.330\ \frac{\text{kJ}}{\text{K}}$.
Substitute into the formula:
$\Delta G = -140 - (327.15 \times -0.330)$
First, calculate $327.15 \times -0.330 \approx -107.96$ (note the negative sign, so multiplying two negatives gives a positive: $327.15 \times 0.330 \approx 107.96$)
$\Delta G = -140 + 107.96 \approx -32.04\ \text{kJ}$ (rounded to zero decimal places: $\Delta G \approx -32\ \text{kJ}$)

Step 4: Determine spontaneity

If $\Delta G < 0$, the reaction is spontaneous. Here, $\Delta G \approx -32\ \text{kJ} < 0$, so this reaction is spontaneous.

Second Reaction: $\boldsymbol{\ce{TiO_2}(s) + 4\ce{HCl}(g)

ightarrow \ce{TiCl_4}(g) + 2\ce{H_2O}(g)}$

Step 1: Use $\boldsymbol{\Delta G = \Delta H - T\Delta S}$ to solve for $\Delta S$

Given $\Delta G = 0\ \text{kJ}$ (at equilibrium), $\Delta H = 70\ \text{kJ}$, and $T = 327.15\ \text{K}$.
Rearrange the formula: $\Delta S = \frac{\Delta H - \Delta G}{T}$
Substitute the values:
$\Delta S = \frac{70 - 0}{327.15} \approx \frac{70}{327.15} \approx 0.214\ \frac{\text{kJ}}{\text{K}}$
Convert to J/K: $0.214\ \frac{\text{kJ}}{\text{K}} \times 1000 = 214\ \frac{\text{J}}{\text{K}}$ (rounded to zero decimal places: $\Delta S \approx 214\ \frac{\text{J}}{\text{K}}$)

Step 2: Determine spontaneity

Since $\Delta G = 0$, the system is at equilibrium, so neither the forward nor reverse reaction is spontaneous (but the problem’s table might have a “neither” option, or since $\Delta G = 0$, it’s at equilibrium).

Final Table Entries:

• First reaction: $\Delta G = \boldsymbol{-32\ \text{kJ}}$, Spontaneous: $\boldsymbol{\text{this reaction}}$
• Second reaction: $\Delta S = \boldsymbol{214\ \frac{\text{J}}{\text{K}}}$, Spontaneous: $\boldsymbol{\text{neither}}$ (or “at equilibrium”)

Answer:

s:

• First reaction $\Delta G$: $\boldsymbol{-32\ \text{kJ}}$, Spontaneous: $\boldsymbol{\text{this reaction}}$
• Second reaction $\Delta S$: $\boldsymbol{214\ \frac{\text{J}}{\text{K}}}$, Spontaneous: $\boldsymbol{\text{neither}}$ (or “at equilibrium”)