Calculate the equilibrium concentration ratio of C to A if equimolar ratio of A and B were allowed to come to equilibrium at 300K.
`A(g)+B(g) iff C(g) +D(g) , DeltaG^(@)=`-830 cal.

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Write the Reaction
We start with the given equilibrium reaction:
\[ A(g) + B(g) \iff C(g) + D(g) \]

### Step 2: Set Initial Concentrations
Since we have an equimolar ratio of A and B, let’s assume the initial concentration of A and B is \( C \) moles each:
- \([A]_{initial} = C\)
- \([B]_{initial} = C\)
- \([C]_{initial} = 0\)
- \([D]_{initial} = 0\)

### Step 3: Define Change in Concentration
Let \( \alpha \) be the fraction of A and B that reacts at equilibrium. Thus, at equilibrium:
- \([A] = C(1 - \alpha)\)
- \([B] = C(1 - \alpha)\)
- \([C] = C\alpha\)
- \([D] = C\alpha\)

### Step 4: Write the Expression for the Equilibrium Constant \( K \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(C\alpha)(C\alpha)}{(C(1 - \alpha))(C(1 - \alpha))} \]
This simplifies to:
\[ K_c = \frac{C^2 \alpha^2}{C^2 (1 - \alpha)^2} = \frac{\alpha^2}{(1 - \alpha)^2} \]

### Step 5: Use the Given \( \Delta G^0 \) to Find \( K_c \)
We know that:
\[ \Delta G^0 = -RT \ln K_c \]
Given \( \Delta G^0 = -830 \) cal, \( R \approx 2 \) cal/(K·mol), and \( T = 300 \) K, we can substitute:
\[ -830 = -2 \times 300 \ln K_c \]
Solving for \( K_c \):
\[ 830 = 600 \ln K_c \]
\[ \ln K_c = \frac{830}{600} \approx 1.3833 \]
Taking the exponential:
\[ K_c = e^{1.3833} \approx 4 \]

### Step 6: Set Up the Equation for \( K_c \)
Now we have:
\[ \frac{\alpha^2}{(1 - \alpha)^2} = 4 \]
Taking the square root of both sides:
\[ \frac{\alpha}{1 - \alpha} = 2 \]

### Step 7: Solve for \( \alpha \)
Cross-multiplying gives:
\[ \alpha = 2(1 - \alpha) \]
\[ \alpha = 2 - 2\alpha \]
\[ 3\alpha = 2 \]
\[ \alpha = \frac{2}{3} \]

### Step 8: Calculate the Ratio of Concentration of C to A
The ratio of the concentration of C to A at equilibrium is:
\[ \frac{[C]}{[A]} = \frac{C\alpha}{C(1 - \alpha)} = \frac{\alpha}{1 - \alpha} \]
Substituting the value of \( \alpha \):
\[ \frac{[C]}{[A]} = \frac{\frac{2}{3}}{1 - \frac{2}{3}} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2 \]

### Final Answer
Thus, the equilibrium concentration ratio of C to A is:
\[ \frac{[C]}{[A]} = 2 \]

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