Calculate `DeltaH_(f)^(@)` for `Ubr_(4)` from the `DeltaG^(@)` of reaction and the `S^(@)` values at 298 K. `U(s)+2Br_(2)(l)rarrUBr_(4)(s),
DeltaG^(@)=-788.6 kJ. " "S^(@)(J//K-"mol")50.3, 152.3, 242.6`

Options

A. `-822.1 kJ//mol`
B. `-841.2 kJ//mol`
C. `-775.6 kJ//mol`
D. `-804.3 kJ//mol`

Explanation

To calculate the standard enthalpy of formation (ΔH_f^(@)) for UBr₄ from the provided ΔG^(@) of the reaction and the standard entropy (S^(@)) values at 298 K, we can follow these steps:

### Step 1: Write the reaction and identify the given values
The reaction is:
\[ \text{U}(s) + 2\text{Br}_2(l) \rightarrow \text{UBr}_4(s) \]

Given:
- ΔG^(@) = -788.6 kJ
- S^(@)(U) = 50.3 J/K·mol
- S^(@)(Br₂) = 152.3 J/K·mol
- S^(@)(UBr₄) = 242.6 J/K·mol

### Step 2: Calculate ΔS for the reaction
Using the formula for the change in entropy (ΔS):
\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]

Substituting the values:
\[ \Delta S = S(UBr₄) - [S(U) + 2 \cdot S(Br₂)] \]
\[ \Delta S = 242.6 - [50.3 + 2 \cdot 152.3] \]
\[ \Delta S = 242.6 - [50.3 + 304.6] \]
\[ \Delta S = 242.6 - 354.9 \]
\[ \Delta S = -112.3 \text{ J/K·mol} \]

### Step 3: Convert ΔS to kJ
Since ΔG is in kJ, we need to convert ΔS from J to kJ:
\[ \Delta S = -112.3 \text{ J/K·mol} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = -0.1123 \text{ kJ/K·mol} \]

### Step 4: Use the Gibbs free energy equation
The Gibbs free energy equation is:
\[ \Delta G = \Delta H - T \Delta S \]

Rearranging to find ΔH:
\[ \Delta H = \Delta G + T \Delta S \]

### Step 5: Substitute the values into the equation
Using T = 298 K:
\[ \Delta H = -788.6 \text{ kJ} + 298 \text{ K} \times (-0.1123 \text{ kJ/K·mol}) \]
\[ \Delta H = -788.6 \text{ kJ} - 33.4714 \text{ kJ} \]
\[ \Delta H = -822.0714 \text{ kJ} \]

### Step 6: Round to appropriate significant figures
Thus, the standard enthalpy of formation for UBr₄ is:
\[ \Delta H_f^(@) \approx -822.1 \text{ kJ/mol} \]

### Final Answer:
\[ \Delta H_f^(@) \text{ for } UBr₄ = -822.1 \text{ kJ/mol} \]

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