The heat of formation of HCl at 348 K from the following data will be :
`0.5H_(2)(g)+0.5Cl_(2)(g)toHCl " "DeltaH_(298)^(@)=-22060` cal. The mean heat capacities over this temperature range are,
`H_(2)(g)," " C_(p)=6.82 " cal" " mol"^(-1)K^(-1)`
`Cl_(2)(g)," " C_(p)=7.71 " cal" " mol"^(-1)K^(-1)`
`HCl(g)," " C_(p)=6.81 " cal" " mol"^(-1)K^(-1)`

Options

A. `-20095` cal
B. `-32758` cal
C. `-37725` cal
D. `-22083` cal

Answer

To find the heat of formation of HCl at 348 K, we will follow these steps:

### Step 1: Calculate ΔCp for the reaction
We need to calculate the change in heat capacity (ΔCp) for the reaction using the formula:

\[ \Delta C_p = C_p(\text{products}) - C_p(\text{reactants}) \]

For the reaction:

\[ 0.5 H_2(g) + 0.5 Cl_2(g) \rightarrow HCl(g) \]

The heat capacities given are:
- \(C_p(HCl) = 6.81 \, \text{cal mol}^{-1} \text{K}^{-1}\)
- \(C_p(H_2) = 6.82 \, \text{cal mol}^{-1} \text{K}^{-1}\)
- \(C_p(Cl_2) = 7.71 \, \text{cal mol}^{-1} \text{K}^{-1}\)

Now substituting these values into the ΔCp formula:

\[ \Delta C_p = C_p(HCl) - \left(0.5 \times C_p(H_2) + 0.5 \times C_p(Cl_2)\right) \]

Calculating the reactants' contribution:

\[ \Delta C_p = 6.81 - \left(0.5 \times 6.82 + 0.5 \times 7.71\right) \]

Calculating the reactants:

\[ = 6.81 - \left(3.41 + 3.855\right) = 6.81 - 7.265 = -0.455 \, \text{cal mol}^{-1} \text{K}^{-1} \]

### Step 2: Calculate ΔH at 348 K
Now we can calculate the heat of formation at 348 K using the formula:

\[ \Delta H(348 \, K) = \Delta H(298 \, K) + \Delta C_p \times \Delta T \]

Where:
- \(\Delta H(298 \, K) = -22060 \, \text{cal}\)
- \(\Delta C_p = -0.455 \, \text{cal mol}^{-1} \text{K}^{-1}\)
- \(\Delta T = 348 \, K - 298 \, K = 50 \, K\)

Substituting these values:

\[ \Delta H(348 \, K) = -22060 + (-0.455) \times 50 \]

Calculating the second term:

\[ = -22060 - 22.75 = -22082.75 \, \text{cal} \]

### Step 3: Rounding the final answer
Rounding to the nearest whole number gives:

\[ \Delta H(348 \, K) \approx -22083 \, \text{cal} \]

### Final Answer
The heat of formation of HCl at 348 K is approximately:

\[ \Delta H(348 \, K) = -22083 \, \text{cal} \]

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