For silver, `C_p(JK^(-1)mol^(-1))=23+0.01T` If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of 'Delta H' will be close to

Options

A. 21 kJ
B. 13 kJ
C. 16 kJ
D. 62 kJ

Explanation

To find the change in enthalpy (ΔH) for the given process, we will use the formula for ΔH at constant pressure, which is given by:

\[ \Delta H = n \int_{T_1}^{T_2} C_p \, dT \]

Where:
- \( n \) = number of moles of the substance
- \( C_p \) = molar heat capacity at constant pressure
- \( T_1 \) = initial temperature
- \( T_2 \) = final temperature

### Step 1: Identify the given values
- \( n = 3 \) moles
- \( T_1 = 300 \, K \)
- \( T_2 = 1000 \, K \)
- \( C_p = 23 + 0.01T \)

### Step 2: Set up the integral
We need to calculate:

\[ \Delta H = n \int_{T_1}^{T_2} C_p \, dT = 3 \int_{300}^{1000} (23 + 0.01T) \, dT \]

### Step 3: Calculate the integral
Now, we can split the integral:

\[ \Delta H = 3 \left( \int_{300}^{1000} 23 \, dT + \int_{300}^{1000} 0.01T \, dT \right) \]

Calculating each part:

1. **First integral:**
\[ \int_{300}^{1000} 23 \, dT = 23[T]_{300}^{1000} = 23(1000 - 300) = 23 \times 700 = 16100 \]

2. **Second integral:**
\[ \int_{300}^{1000} 0.01T \, dT = 0.01 \left[ \frac{T^2}{2} \right]_{300}^{1000} = 0.01 \left( \frac{1000^2}{2} - \frac{300^2}{2} \right) \]
\[ = 0.01 \left( \frac{1000000}{2} - \frac{90000}{2} \right) = 0.01 \left( 500000 - 45000 \right) = 0.01 \times 455000 = 4550 \]

### Step 4: Combine the results
Now, substituting back into the ΔH equation:

\[ \Delta H = 3 \left( 16100 + 4550 \right) = 3 \times 20550 = 61650 \, J \]

### Step 5: Convert to kilojoules
To convert joules to kilojoules:

\[ \Delta H = \frac{61650}{1000} = 61.65 \, kJ \]

### Step 6: Round off the answer
Rounding to the nearest whole number gives:

\[ \Delta H \approx 62 \, kJ \]

Thus, the value of ΔH will be close to **62 kJ**.

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