The reaction `A+2B+C to Z` occurs by the following mechanism:
(i) `A+2B+C to X` (fast)
(ii) `X+C to Y` (slow) (iii) `Y+B to Z` (very fast)
Rate law for this reaction is:

Options

A. k [C]
B. `k[A] [B]^2[C]^2
C. `k[Z]`
D. `k[X][Y][Z]`

Answer

To determine the rate law for the reaction \( A + 2B + C \to Z \) given the mechanism, we will analyze each step of the mechanism and derive the rate law based on the slow step, which is the rate-determining step.

### Step-by-Step Solution:

1. **Identify the Mechanism Steps**:
- Step (i): \( A + 2B + C \to X \) (fast)
- Step (ii): \( X + C \to Y \) (slow)
- Step (iii): \( Y + B \to Z \) (very fast)

2. **Determine the Rate-Determining Step**:
- The slow step (ii) is the rate-determining step. The rate of the overall reaction is determined by this step.

3. **Write the Rate Law for the Slow Step**:
- For the slow step \( X + C \to Y \), the rate law can be expressed as:
\[ \text{Rate} = k [X][C] \]
where \( k \) is the rate constant for the slow step.

4. **Express \( [X] \) in Terms of Reactants**:
- Since \( X \) is an intermediate and does not appear in the overall reaction, we need to express \( [X] \) in terms of the reactants \( A \), \( B \), and \( C \).
- From the fast step (i), we can assume that it reaches a quasi-equilibrium condition due to its fast nature. Therefore, we can write:
\[ [X] = k_{eq} [A][B]^2[C] \]
where \( k_{eq} \) is the equilibrium constant for the fast step.

5. **Substitute \( [X] \) into the Rate Law**:
- Now, substituting \( [X] \) back into the rate law derived from the slow step:
\[ \text{Rate} = k [X][C] = k (k_{eq} [A][B]^2[C])[C] \]
- Simplifying this gives:
\[ \text{Rate} = k k_{eq} [A][B]^2[C]^2 \]

6. **Final Rate Law Expression**:
- We can denote \( k' = k k_{eq} \) (a new rate constant) to simplify the expression:
\[ \text{Rate} = k' [A][B]^2[C]^2 \]

### Conclusion:
The rate law for the reaction \( A + 2B + C \to Z \) is:
\[ \text{Rate} = k' [A][B]^2[C]^2 \]