For a reaction
`aA(g)hArrbB(g)`
at equilibrium, the heat of reaction at constant volume is `1500` cal more than at constant pressure. If the temperature is `27^(@)C` then

Options

A. `K_(p)=K_(c )`
B. `K_(p) gt K_(c )`
C. `K_(p) lt K_(c )`
D. None of these

Explanation

To solve the problem step by step, we need to analyze the given information and apply the relevant concepts from chemical equilibrium.

### Step 1: Understand the Relationship Between Enthalpy and Internal Energy
For a reaction at equilibrium, the relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) at constant pressure and constant volume is given by:
\[ \Delta H = \Delta U + P \Delta V \]
At constant volume, ΔH equals ΔU because there is no volume change (ΔV = 0).

### Step 2: Analyze the Given Information
We are given that the heat of reaction at constant volume is 1500 cal more than at constant pressure:
\[ \Delta H_{V} = \Delta H_{P} + 1500 \text{ cal} \]
From the relationship established in Step 1, we can substitute:
\[ \Delta U = \Delta H_{P} - P \Delta V \]
Thus, we can express the relationship as:
\[ \Delta H_{V} = \Delta U + 1500 \text{ cal} \]

### Step 3: Relate ΔH and ΔU
Since ΔH at constant volume equals ΔU, we can write:
\[ \Delta U + 1500 = \Delta H_{P} \]
This implies:
\[ \Delta H_{P} - \Delta U = 1500 \text{ cal} \]

### Step 4: Consider the Change in Moles of Gas (Δn)
The difference between ΔH and ΔU can be expressed in terms of the change in the number of moles of gas (Δn):
\[ \Delta H - \Delta U = \Delta n \cdot R \cdot T \]
Where R is the gas constant and T is the temperature in Kelvin.

### Step 5: Calculate Temperature in Kelvin
Given that the temperature is 27°C, we convert it to Kelvin:
\[ T = 27 + 273 = 300 \text{ K} \]

### Step 6: Substitute Values
Now we can substitute the known values into the equation:
\[ 1500 \text{ cal} = \Delta n \cdot R \cdot 300 \text{ K} \]
Using R = 1.987 cal/(mol·K), we can find Δn:
\[ 1500 = \Delta n \cdot 1.987 \cdot 300 \]
\[ \Delta n = \frac{1500}{1.987 \cdot 300} \]

### Step 7: Calculate Δn
Calculating Δn gives:
\[ \Delta n = \frac{1500}{596.1} \approx 2.52 \]
This indicates that the change in moles of gas is approximately 2.52.

### Step 8: Determine Kp and Kc Relationship
From the previous analysis, we know that:
\[ K_p = K_c (RT)^{\Delta n} \]
Since Δn is positive, it implies that Kp is greater than Kc:
\[ K_p > K_c \]

### Conclusion
Thus, we conclude that the relationship between Kp and Kc is:
\[ K_p > K_c \]