Find out the Value ofequilibrium constant for the following reaction at 298 K, 2 NH3(g) + CO2 ⇌ NH2CONH2(aq) + H2O(I) Standard Gibbs energy change, AGr° at the given temperature is – 13.6 kJ mol-1.

Answer

To find the value of the equilibrium constant (K) for the reaction:

\[ 2 \, \text{NH}_3(g) + \text{CO}_2 \rightleftharpoons \text{NH}_2\text{CONH}_2(aq) + \text{H}_2\text{O}(l) \]

given that the standard Gibbs energy change (\( \Delta G^\circ \)) at 298 K is -13.6 kJ/mol, we can follow these steps:

### Step 1: Convert Gibbs Free Energy Change to Joules
Since the Gibbs free energy change is given in kJ/mol, we need to convert it into Joules/mol for consistency with the gas constant (R).

\[ \Delta G^\circ = -13.6 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -13600 \, \text{J/mol} \]

### Step 2: Use the Relationship Between Gibbs Free Energy and Equilibrium Constant
The relationship between the standard Gibbs free energy change and the equilibrium constant (K) is given by the equation:

\[ \Delta G^\circ = -RT \ln K \]

Where:
- \( R \) is the universal gas constant (8.314 J/(mol·K))
- \( T \) is the temperature in Kelvin (298 K)

### Step 3: Rearrange the Equation to Solve for K
We can rearrange the equation to solve for K:

\[ \ln K = -\frac{\Delta G^\circ}{RT} \]

### Step 4: Substitute the Values into the Equation
Now, we substitute the values we have into the equation:

\[ \ln K = -\frac{-13600 \, \text{J/mol}}{(8.314 \, \text{J/(mol·K)}) \times (298 \, \text{K})} \]

### Step 5: Calculate the Right Side
Calculating the denominator:

\[ RT = 8.314 \, \text{J/(mol·K)} \times 298 \, \text{K} = 2477.572 \, \text{J/mol} \]

Now substituting back:

\[ \ln K = \frac{13600}{2477.572} \approx 5.49 \]

### Step 6: Convert from Natural Log to Base 10 Log
To find K, we need to convert from natural logarithm to base 10 logarithm:

\[ \ln K = 5.49 \implies \log_{10} K = \frac{5.49}{2.303} \approx 2.38 \]

### Step 7: Calculate K using Antilog
Now, we can calculate K by taking the antilog:

\[ K = 10^{2.38} \approx 239.88 \]

### Final Answer
Thus, the value of the equilibrium constant \( K \) for the reaction at 298 K is approximately:

\[ K \approx 239.88 \]