Given the following reaction and equilibrium constant
`CO(g)+(1)/(2)O_(2)(g)hArrCO_(2)(g),K_(c)=1.1xx10^(11)`
So at equilibrium select the correct response from below:-

Options

A. The net driving force for the reaction will always be left to right because the equilibrium constant is very large
B. The rate of forward reaction will always be greater than the rate of the reverse reaction because the equilibrium constant is so large
C. For a reaction mixture containing equilibrium concentrations of the three components, the forward and reverse rates will both go to zero
D. All responses above are incorrect

Answer

To solve the problem regarding the equilibrium constant and the behavior of the reaction at equilibrium, we can follow these steps:

### Step 1: Understand the Reaction and Equilibrium Constant
The reaction given is:
\[ \text{CO(g)} + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{CO}_2(g) \]
The equilibrium constant \( K_c \) for this reaction is given as \( 1.1 \times 10^{11} \).

### Step 2: Analyze the Value of the Equilibrium Constant
A large value of \( K_c \) (in this case, \( 1.1 \times 10^{11} \)) indicates that at equilibrium, the products are favored over the reactants. This means that the reaction proceeds almost to completion.

### Step 3: Understand the Concept of Dynamic Equilibrium
At equilibrium, the rates of the forward and reverse reactions are equal. This means that the concentration of reactants and products remains constant over time, even though both reactions are still occurring.

### Step 4: Evaluate the Statements
Now, we need to evaluate the correctness of the statements provided in the question:

1. **Statement 1**: "The net driving force for the reaction will always be left to right because the equilibrium constant is very large."
- This statement is incorrect because, at equilibrium, the driving force is balanced; the rates of the forward and reverse reactions are equal.

2. **Statement 2**: "The rate of forward reaction will always be greater than the rate of reverse reaction because the equilibrium constant is so large."
- This statement is also incorrect. At equilibrium, the rates of the forward and reverse reactions are equal, regardless of the value of \( K_c \).

3. **Statement 3**: "For the reaction mixture containing equilibrium concentrations of the three components, the forward and reverse rates will both go to zero."
- This statement is incorrect as well. The rates do not go to zero; they remain equal and non-zero at equilibrium.

4. **Statement 4**: "All responses above are incorrect."
- This statement is correct because all previous statements have been shown to be incorrect.

### Conclusion
Based on the analysis, the correct answer is:
**D) All responses above are incorrect.**