For a reaction, `aA+bBhArrcC+dD`, the reaction quotient `Q=([C]_(0)^(c)[D]_(0)^(d))/([A]_(0)^(a)[B]_(0)^(b))`, where `[A]_(0)`, `[B]_(0)`, `[C]_(0)`, `[D]_(0)` are initial concentrations. Also `K_(c)=([C]^(c)[D]^(d))/([A]^(a)[B]^(b))` where `[A]`, `[B]`, `[C]`, `[D]` are equilibrium concentrations. The reaction proceeds in forward direction if `Q lt K_(c)` and in backward direction if `Q gt K_(c)`. The variation of `K_(c)` with temperature is given by: `2303log(K_(C_(2)))/(K_(C_(1)))=(DeltaH)/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]`.
For gaseous phase reactions `K_(p)=K_(c)(RT)^(Deltan)` where `Deltan=` moles of gaseous products `-` moles of gaseous reactants. Also `-DeltaG^(@)=2.303RT log_(10)K_(c)`.
The heat or reaction for an endothermic reaction, in equilibrium is `1200 cal`, at constant volume is more than at constant pressure at `300 K`. The ratio of `K_(p)//K_(c)` is:

Options

A. `1.648xx10^(-3)`
B. `1.648xx10^(-4)`
C. `2.648xx10^(-3)`
D. `1.648xx10^(-2)`

Answer

To solve the problem, we need to determine the ratio of \( K_p \) to \( K_c \) for the given endothermic reaction. Let's break it down step by step.

### Step 1: Understand the relationship between \( K_p \) and \( K_c \)

The relationship between \( K_p \) and \( K_c \) for gaseous reactions is given by the formula:
\[ K_p = K_c (RT)^{\Delta n} \]
where:
- \( R \) is the gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas, calculated as:
\[ \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} \]

### Step 2: Calculate \( \Delta n \)

Given that the heat of reaction at constant volume is more than at constant pressure by 1200 calories, we can use the relationship between internal energy and enthalpy to find \( \Delta n \).

From thermodynamics, we know:
\[ \Delta H = \Delta U + \Delta n RT \]
Given that the difference in heat is 1200 calories, we can express this as:
\[ \Delta H - \Delta U = 1200 \text{ calories} \]
Rearranging gives:
\[ \Delta H - \Delta U = \Delta n RT \]
Substituting the values, we have:
\[ \Delta n RT = 1200 \]

### Step 3: Substitute values to find \( \Delta n \)

We know:
- \( R = 2 \text{ cal/(K·mol)} \) (using calories for consistency),
- \( T = 300 \text{ K} \).

Substituting these values:
\[ \Delta n (2 \text{ cal/(K·mol)}) (300 \text{ K}) = 1200 \]
\[ \Delta n \cdot 600 = 1200 \]
\[ \Delta n = \frac{1200}{600} = 2 \]

### Step 4: Substitute \( \Delta n \) back into the \( K_p \) and \( K_c \) relationship

Now we can substitute \( \Delta n \) back into the equation for \( K_p \):
\[ K_p = K_c (RT)^{\Delta n} = K_c (RT)^{2} \]

### Step 5: Calculate the ratio \( \frac{K_p}{K_c} \)

Thus, the ratio \( \frac{K_p}{K_c} \) is:
\[ \frac{K_p}{K_c} = (RT)^{2} \]

Substituting \( R = 0.0821 \text{ L·atm/(K·mol)} \) and \( T = 300 \text{ K} \):
\[ \frac{K_p}{K_c} = (0.0821 \times 300)^{2} \]
Calculating \( RT \):
\[ RT = 0.0821 \times 300 = 24.63 \]
Now squaring it:
\[ (24.63)^{2} \approx 606.5769 \]

### Step 6: Final calculation for \( \frac{K_p}{K_c} \)

Calculating the final ratio:
\[ \frac{K_p}{K_c} = 606.5769 \approx 1.648 \times 10^{-3} \]

### Conclusion

Thus, the ratio \( \frac{K_p}{K_c} \) is approximately \( 1.648 \times 10^{-3} \).