Question

using the thermodynamic information in the aleks data tab, calculate the standard reaction entropy of the following chemical reaction:
$2\ce{nh_3}(g) \
ightarrow \ce{n_2h_4}(g) + \ce{h_2}(g)$
round your answer to zero decimal places.

Explanation:

Step1: Recall the formula for standard reaction entropy

The formula for the standard reaction entropy ($\Delta S^{\circ}$) is $\Delta S^{\circ}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients, and $S^{\circ}$ is the standard molar entropy.

Step2: Find the standard molar entropies (from ALEKS Data tab)

From thermodynamic data:
- $S^{\circ}(\text{NH}_3(g)) = 192.8\ \frac{\text{J}}{\text{mol·K}}$
- $S^{\circ}(\text{N}_2\text{H}_4(g)) = 238.5\ \frac{\text{J}}{\text{mol·K}}$
- $S^{\circ}(\text{H}_2(g)) = 130.7\ \frac{\text{J}}{\text{mol·K}}$

Step3: Calculate the entropy of products

For products: $\text{N}_2\text{H}_4(g)$ and $\text{H}_2(g)$. The stoichiometric coefficients are 1 for $\text{N}_2\text{H}_4$ and 1 for $\text{H}_2$.
$\sum nS^{\circ}(\text{products})=1\times S^{\circ}(\text{N}_2\text{H}_4)+1\times S^{\circ}(\text{H}_2)=238.5 + 130.7=369.2\ \frac{\text{J}}{\text{K}}$ (since moles: 1 mol of $\text{N}_2\text{H}_4$ and 1 mol of $\text{H}_2$, so total entropy of products is sum of their entropies)

Step4: Calculate the entropy of reactants

For reactant: $\text{NH}_3(g)$ with stoichiometric coefficient 2.
$\sum mS^{\circ}(\text{reactants})=2\times S^{\circ}(\text{NH}_3)=2\times192.8 = 385.6\ \frac{\text{J}}{\text{K}}$

Step5: Calculate $\Delta S^{\circ}$

$\Delta S^{\circ}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})=369.2-385.6=- 16.4\ \frac{\text{J}}{\text{K}}$? Wait, no, wait. Wait, the reaction is $2\text{NH}_3(g)
ightarrow\text{N}_2\text{H}_4(g)+\text{H}_2(g)$. So products are 1 mol $\text{N}_2\text{H}_4$ and 1 mol $\text{H}_2$, reactants are 2 mol $\text{NH}_3$. Wait, let's recalculate:

Wait, $S^{\circ}(\text{NH}_3) = 192.8\ \text{J/(mol·K)}$, so 2 mol: $2\times192.8 = 385.6\ \text{J/K}$ (entropy of reactants)

Products: 1 mol $\text{N}_2\text{H}_4$: $238.5\ \text{J/K}$, 1 mol $\text{H}_2$: $130.7\ \text{J/K}$. So total products entropy: $238.5 + 130.7=369.2\ \text{J/K}$

Then $\Delta S^{\circ}=369.2 - 385.6=- 16.4\ \text{J/K}$? But wait, maybe I made a mistake in data? Wait, let's check the standard entropies again. Wait, maybe the values are different? Wait, let's confirm the standard entropies:

Wait, actual standard molar entropies (from reliable sources):
- $\text{NH}_3(g)$: $192.8\ \text{J/(mol·K)}$
- $\text{N}_2\text{H}_4(g)$: $238.5\ \text{J/(mol·K)}$
- $\text{H}_2(g)$: $130.684\ \text{J/(mol·K)}$ (more accurate value is 130.684)

Let's recalculate with more accurate $S^{\circ}(\text{H}_2)=130.684$:

Products: $238.5+130.684 = 369.184\ \text{J/K}$

Reactants: $2\times192.8 = 385.6\ \text{J/K}$

$\Delta S^{\circ}=369.184 - 385.6=- 16.416\ \text{J/K}$, which rounds to - 16 $\text{J/K}$? Wait, but maybe I had the formula reversed? Wait, no: $\Delta S^{\circ}=\sum S^{\circ}(\text{products})-\sum S^{\circ}(\text{reactants})$. Wait, but let's check the stoichiometry again. The reaction is $2\text{NH}_3
ightarrow\text{N}_2\text{H}_4+\text{H}_2$. So moles of reactants: 2 mol $\text{NH}_3$, moles of products: 1 mol $\text{N}_2\text{H}_4$ and 1 mol $\text{H}_2$. So the formula is correct.

Wait, maybe the standard entropies are different? Let's check another source. For $\text{NH}_3(g)$: $192.45\ \text{J/(mol·K)}$, $\text{N}_2\text{H}_4(g)$: $238.36\ \text{J/(mol·K)}$, $\text{H}_2(g)$: $130.684\ \text{J/(mol·K)}$.

Then products: $238.36 + 130.684=369.044\ \text{J/K}$

Reactants: $2\times192.45 = 384.9\ \text{J/K}$

$\Delta S^{\circ}=369.044 - 384.9=- 15.856\approx - 16\ \text{J/K}$ (rounded to zero decimal places)

Wait, but maybe I made a mistake in the sign? Wait, no, the number of moles: reactants are 2 moles of gas, products are 1 + 1 = 2 moles of gas. Wait, 2 moles of reactant gas (NH3) and 2 moles of product gas (N2H4 + H2). Wait, 2 moles reactant, 2 moles product. So the change in moles of gas is zero. But the entropy values: NH3 has S=192.45, N2H4 has 238.36, H2 has 130.684. So 2*192.45 = 384.9; 238.36 + 130.684 = 369.044. So 369.044 - 384.9 = -15.856, which is approximately -16 J/K.

Answer:

-16