For the reaction , `H_(2)(g)+1//2O_(2)(g)=H_(2)O(l), Delta C_(p)=7.63 "cal/deg" , Delta H_(25^(@)C)=68.3` Kcal, what will be the value (in Kcal) of `Delta H` at `100^(@)C` :

Options

A. `7.63xx(373-298)-68.3`
B. `7.63xx10^(-3)(373-298)-68.3`
C. `7.63xx10^(-3)(373-298)+68.3`
D. `7.63xx(373-298)+68.3`

Explanation

To find the value of ΔH at 100°C for the given reaction, we can use the following formula:

\[ \Delta H(T) = \Delta H(T_1) + \Delta C_p \times (T - T_1) \]

where:
- \( \Delta H(T) \) is the enthalpy change at temperature \( T \),
- \( \Delta H(T_1) \) is the enthalpy change at the reference temperature \( T_1 \),
- \( \Delta C_p \) is the heat capacity change,
- \( T \) is the final temperature,
- \( T_1 \) is the initial temperature.

### Step 1: Identify the given values
- \( \Delta H(25°C) = 68.3 \, \text{kcal} \)
- \( \Delta C_p = 7.63 \, \text{cal/°C} \)
- Convert \( \Delta C_p \) to kcal:
\[ \Delta C_p = 7.63 \, \text{cal/°C} \times \frac{1 \, \text{kcal}}{1000 \, \text{cal}} = 0.00763 \, \text{kcal/°C} \]
- \( T_1 = 25°C \)
- \( T = 100°C \)

### Step 2: Calculate the change in temperature
\[ \Delta T = T - T_1 = 100°C - 25°C = 75°C \]

### Step 3: Substitute the values into the formula
\[ \Delta H(100°C) = \Delta H(25°C) + \Delta C_p \times \Delta T \]
\[ \Delta H(100°C) = 68.3 \, \text{kcal} + 0.00763 \, \text{kcal/°C} \times 75°C \]

### Step 4: Calculate the additional enthalpy change
\[ \Delta H(100°C) = 68.3 \, \text{kcal} + 0.00763 \times 75 \]
\[ \Delta H(100°C) = 68.3 \, \text{kcal} + 0.57225 \, \text{kcal} \]
\[ \Delta H(100°C) = 68.87225 \, \text{kcal} \]

### Step 5: Final answer
Rounding to three significant figures, we get:
\[ \Delta H(100°C) \approx 68.87 \, \text{kcal} \]