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 reaction \( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \), we can use the following steps:

### Step 1: Identify the given data
- ΔCp = 7.63 cal/°C
- ΔH at 25°C (ΔH1) = 68.3 kcal
- T1 = 25°C
- T2 = 100°C

### Step 2: Convert ΔCp to kcal/°C
Since we need to express ΔCp in kcal, we convert it as follows:
\[ \Delta Cp = 7.63 \text{ cal/°C} = 7.63 \times 10^{-3} \text{ kcal/°C} \]

### Step 3: Convert temperatures from °C to K
- T1 in Kelvin = 25 + 273 = 298 K
- T2 in Kelvin = 100 + 273 = 373 K

### Step 4: Calculate the change in temperature (ΔT)
\[ \Delta T = T2 - T1 = 373 \text{ K} - 298 \text{ K} = 75 \text{ K} \]

### Step 5: Use the formula for ΔH
The change in enthalpy (ΔH) can be calculated using the equation:
\[ \Delta H = \Delta H1 + \Delta Cp \times \Delta T \]
Substituting the known values:
\[ \Delta H2 = 68.3 \text{ kcal} + (7.63 \times 10^{-3} \text{ kcal/°C}) \times 75 \text{ °C} \]

### Step 6: Calculate ΔH2
Calculating the second term:
\[ \Delta Cp \times \Delta T = 7.63 \times 10^{-3} \times 75 = 0.57225 \text{ kcal} \]
Now substituting this back into the equation:
\[ \Delta H2 = 68.3 \text{ kcal} + 0.57225 \text{ kcal} = 68.87225 \text{ kcal} \]

### Step 7: Round off the final answer
Rounding to two decimal places, we find:
\[ \Delta H2 \approx 68.87 \text{ kcal} \]

### Final Answer
The value of ΔH at 100°C is approximately **68.87 kcal**.

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