For the reaction , `H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)(l),DeltaC_(p)=32 JK^(-1), DeltaH ` at `27^(@)C = - 285.8 kJ mol^(-1). What will be the value of `Delta H ` at `127@degreeC` ?

Options

A. `- 289 . 0 kJ mol^(-1)`
B. ` - 282 . 6 kJ mol^(-1)`
C. ` -317 .0 kJ mol^(-1)`
D. ` -253.8 kJ mol^(-1)`

Explanation

To find the value of ΔH at 127°C for the reaction \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2(l) \), we can use Kirchhoff's equation, which relates the change in enthalpy with temperature. The equation is given by:

\[ \Delta H_2 - \Delta H_1 = \Delta C_p \times (T_2 - T_1) \]

Where:
- \( \Delta H_1 \) is the enthalpy change at the initial temperature \( T_1 \).
- \( \Delta H_2 \) is the enthalpy change at the final temperature \( T_2 \).
- \( \Delta C_p \) is the change in heat capacity.
- \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.

### Step-by-step Solution:

1. **Convert temperatures from Celsius to Kelvin:**
- \( T_1 = 27°C = 27 + 273.15 = 300.15 K \)
- \( T_2 = 127°C = 127 + 273.15 = 400.15 K \)

2. **Identify the given values:**
- \( \Delta H_1 = -285.8 \, \text{kJ/mol} \)
- \( \Delta C_p = 32 \, \text{J/K} = 0.032 \, \text{kJ/K} \) (since 1 kJ = 1000 J)

3. **Calculate the temperature difference:**
- \( \Delta T = T_2 - T_1 = 400.15 K - 300.15 K = 100 K \)

4. **Apply Kirchhoff's equation:**
\[ \Delta H_2 - \Delta H_1 = \Delta C_p \times \Delta T \]
\[ \Delta H_2 - (-285.8) = 0.032 \, \text{kJ/K} \times 100 \, \text{K} \]
\[ \Delta H_2 + 285.8 = 3.2 \, \text{kJ} \]

5. **Solve for \( \Delta H_2 \):**
\[ \Delta H_2 = 3.2 - 285.8 \]
\[ \Delta H_2 = -282.6 \, \text{kJ/mol} \]

### Final Answer:
The value of \( \Delta H \) at \( 127°C \) is \( -282.6 \, \text{kJ/mol} \).

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