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using the thermodynamic information in the aleks data tab, calculate th…

Question

using the thermodynamic information in the aleks data tab, calculate the standard reaction entropy of the following chemical reaction: 2nh₃(g) → n₂h₄(g) + h₂(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 of products and reactants respectively, and $S^{\circ}$ is the standard molar entropy.

Step2: Identify the standard molar entropies (from ALEKS Data tab, typical values: $S^{\circ}(\text{NH}_3(g)) = 192.8\ \text{J/(mol·K)}$, $S^{\circ}(\text{N}_2\text{H}_4(g)) = 238.5\ \text{J/(mol·K)}$, $S^{\circ}(\text{H}_2(g)) = 130.7\ \text{J/(mol·K)}$)

Step3: Calculate the entropy of products

For products: $\text{N}_2\text{H}_4(g)$ has a coefficient of 1, $\text{H}_2(g)$ has a coefficient of 1. So $\sum nS^{\circ}(\text{products})=1\times S^{\circ}(\text{N}_2\text{H}_4(g)) + 1\times S^{\circ}(\text{H}_2(g))=238.5 + 130.7 = 369.2\ \text{J/K}$ (per 2 moles of $\text{NH}_3$? Wait, no, stoichiometry: reactant is $2\text{NH}_3(g)$, so moles of reactant: 2. Products: 1 $\text{N}_2\text{H}_4$ and 1 $\text{H}_2$. So $\sum mS^{\circ}(\text{reactants}) = 2\times S^{\circ}(\text{NH}_3(g))=2\times192.8 = 385.6\ \text{J/K}$

Step4: Calculate $\Delta S^{\circ}$

$\Delta S^{\circ}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})=(238.5 + 130.7)-2\times192.8$
First, calculate products: $238.5 + 130.7 = 369.2$
Reactants: $2\times192.8 = 385.6$
Then $\Delta S^{\circ}=369.2 - 385.6=-16.4\ \text{J/K}$? Wait, no, wait, maybe I mixed up. Wait, let's check again. Wait, the reaction is $2\text{NH}_3(g)
ightarrow\text{N}_2\text{H}_4(g)+\text{H}_2(g)$

So products: 1 mol $\text{N}_2\text{H}_4$ and 1 mol $\text{H}_2$. Reactants: 2 mol $\text{NH}_3$.

So $\sum nS^{\circ}(\text{products}) = S^{\circ}(\text{N}_2\text{H}_4) + S^{\circ}(\text{H}_2)=238.5 + 130.7 = 369.2\ \text{J/K}$ (for 1 mol $\text{N}_2\text{H}_4$ and 1 mol $\text{H}_2$, which is the amount from 2 mol $\text{NH}_3$)

$\sum mS^{\circ}(\text{reactants}) = 2\times S^{\circ}(\text{NH}_3)=2\times192.8 = 385.6\ \text{J/K}$

Then $\Delta S^{\circ}=369.2 - 385.6=-16.4\ \text{J/K}$? But wait, maybe the values from ALEKS are different. Wait, maybe I used wrong values. Let's check correct values (from standard tables or ALEKS):

Wait, actually, correct standard molar entropies (from reliable sources):

$S^{\circ}(\text{NH}_3(g)) = 192.45\ \text{J/(mol·K)}$

$S^{\circ}(\text{N}_2\text{H}_4(g)) = 238.36\ \text{J/(mol·K)}$

$S^{\circ}(\text{H}_2(g)) = 130.684\ \text{J/(mol·K)}$

So recalculating:

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

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

$\Delta S^{\circ}=369.044 - 384.9=-15.856\ \text{J/K}$, which rounds to -16 J/K? Wait, but maybe the ALEKS data has different values. Wait, perhaps I made a mistake in stoichiometry. Wait, the reaction is 2 moles of NH3 producing 1 mole of N2H4 and 1 mole of H2. So the formula is correct.

Wait, maybe the user's ALEKS data has:

$S^{\circ}(\text{NH}_3) = 192.8\ \text{J/(mol·K)}$, $S^{\circ}(\text{N}_2\text{H}_4)=238.5\ \text{J/(mol·K)}$, $S^{\circ}(\text{H}_2)=130.7\ \text{J/(mol·K)}$

Then:

Products: $238.5 + 130.7 = 369.2$

Reactants: $2\times192.8 = 385.6$

$\Delta S^{\circ}=369.2 - 385.6 = -16.4\ \text{J/K}$, which rounds to -16 J/K (to zero decimal places). But wait, maybe the actual values from ALEKS are such that:

Wait, let's check another way. Maybe the standard entropies are:

$\text{NH}_3(g)$: 192.5 J/(mol·K)

$\text{N}_2\text{H}_4(g)$: 238.4 J/(mol·K)

$\text{H}_2(g)$: 130.6 J/(mol·K)

Then products: 238.4 + 130.6 = 369.0

Re…

Answer:

-16 (assuming the calculation with standard values rounds to -16, but actual value depends on ALEKS data; if using the typical values, the answer is -16 J/K)