The heat change for the following reaction at `298 K` and constant pressure is `+7.3 kcal`
`A_(@)B(s)to2A(s)+(1)/(2)B_(2)(g)` ,` DeltaH=+7.3 kcal`
The heat change at constant volume would be

Options

A. `+7.3" Kcal"`
B. more than 7.3 Kcal
C. less than 7.3Kcal
D. zero

Answer

To find the heat change at constant volume for the given reaction, we will use the relationship between enthalpy change (ΔH) and internal energy change (ΔE). The equation we will use is:

\[ \Delta E = \Delta H - \Delta N_g RT \]

Where:
- ΔE = change in internal energy
- ΔH = change in enthalpy (given as +7.3 kcal)
- ΔN_g = change in the number of moles of gaseous products minus gaseous reactants
- R = gas constant (0.001987 kcal/(mol·K))
- T = temperature in Kelvin (given as 298 K)

### Step 1: Calculate ΔN_g
Identify the number of moles of gaseous products and reactants in the reaction:

\[ A_{(s)} + B_{(s)} \rightarrow 2A_{(s)} + \frac{1}{2}B_{2(g)} \]

- Gaseous products: \(\frac{1}{2} B_2\) (which is 0.5 moles)
- Gaseous reactants: 0 (since both A and B are solids)

Thus,

\[ \Delta N_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 0.5 - 0 = 0.5 \]

### Step 2: Substitute values into the equation
Now, substitute ΔH, ΔN_g, R, and T into the equation:

\[ \Delta E = 7.3 \, \text{kcal} - (0.5) \times (0.001987 \, \text{kcal/(mol·K)}) \times (298 \, \text{K}) \]

### Step 3: Calculate the term ΔN_g RT
First, calculate \(0.5 \times 0.001987 \times 298\):

\[ 0.5 \times 0.001987 \times 298 = 0.5 \times 0.592746 = 0.296373 \, \text{kcal} \]

### Step 4: Calculate ΔE
Now substitute this value back into the equation for ΔE:

\[ \Delta E = 7.3 \, \text{kcal} - 0.296373 \, \text{kcal} = 7.003627 \, \text{kcal} \]

### Step 5: Round the value
Rounding this to three decimal places gives:

\[ \Delta E \approx 7.004 \, \text{kcal} \]

Thus, the heat change at constant volume is approximately **7.004 kcal**.