The equilibrium constant of a reaction at 298 K is `5 xx 10^(-3)` and at 1000 K is `2 xx 10^(-5)` What is the sign of `triangleH` for the reaction.

Options

A. `triangleH=0`
B. `triangleH` is negative
C. `triangleH` is positive
D. None of these

Answer

To determine the sign of ΔH (enthalpy change) for the reaction given the equilibrium constants at two different temperatures, we can use the van 't Hoff equation, which relates the change in the equilibrium constant (K) with temperature (T) and the enthalpy change (ΔH) of the reaction.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Equilibrium constant at T1 (298 K): K1 = 5 x 10^(-3)
- Equilibrium constant at T2 (1000 K): K2 = 2 x 10^(-5)

2. **Write the van 't Hoff Equation:**
\[ \ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
where R is the universal gas constant (approximately 8.314 J/(mol·K)).

3. **Calculate the ratio of K2 to K1:**
\[ \frac{K_2}{K_1} = \frac{2 \times 10^{-5}}{5 \times 10^{-3}} = \frac{2}{5} \times 10^{-2} = 0.4 \times 10^{-2} \]

4. **Take the natural logarithm:**
\[ \ln(0.4 \times 10^{-2}) = \ln(0.4) + \ln(10^{-2}) = \ln(0.4) - 2 \ln(10) \]
(Since \(\ln(10) \approx 2.303\), we can calculate \(\ln(0.4)\) using a calculator or logarithm table, which is approximately -0.916.)

Therefore:
\[ \ln(0.4 \times 10^{-2}) \approx -0.916 - 4.606 = -5.522 \]

5. **Calculate the temperature difference:**
\[ \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{1000} - \frac{1}{298} \approx 0.001 - 0.003356 = -0.002356 \text{ K}^{-1} \]

6. **Substitute values into the van 't Hoff equation:**
\[ -5.522 = -\frac{\Delta H}{8.314} \times (-0.002356) \]

7. **Rearranging to find ΔH:**
\[ \Delta H = 5.522 \times 8.314 \times 0.002356 \]

Calculate the value:
\[ \Delta H \approx 5.522 \times 8.314 \times 0.002356 \approx 0.103 \text{ kJ/mol} \]

8. **Determine the sign of ΔH:**
Since K2 < K1, it indicates that the reaction favors the reactants at higher temperature, suggesting that the reaction is exothermic. Therefore, ΔH is negative.

### Conclusion:
The sign of ΔH for the reaction is negative.

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