The rate law for the reaction `2C+D to A+E` is `(-d[D])/(dt)=k[C]^2[D]`. IF C is present in large excess, the order of the reaction will be:

Options

A. Zero order
B. First order
C. Second order
D. Third order

Answer

To determine the order of the reaction given the rate law and the condition that C is present in large excess, we can follow these steps:

### Step 1: Write down the given rate law
The rate law for the reaction is given as:
\[ -\frac{d[D]}{dt} = k[C]^2[D] \]

### Step 2: Analyze the condition of excess reactant
Since C is present in large excess, its concentration can be considered constant during the reaction. This means that the change in concentration of C is negligible compared to D.

### Step 3: Simplify the rate law
When C is in large excess, we can treat \([C]\) as a constant. Let's denote this constant concentration of C as \( [C]_0 \). The rate law can then be rewritten as:
\[ -\frac{d[D]}{dt} = k[C]_0^2[D] \]
Here, \( k[C]_0^2 \) is a new constant, which we can denote as \( k' \):
\[ -\frac{d[D]}{dt} = k'[D] \]

### Step 4: Identify the order of the reaction
The simplified rate law \( -\frac{d[D]}{dt} = k'[D] \) indicates that the rate of the reaction depends only on the concentration of D. This is a first-order reaction with respect to D.

### Step 5: Conclusion
Since the overall rate of the reaction is now dependent solely on the concentration of D, and since we have treated C as a constant, the overall order of the reaction is:
\[ \text{Order of the reaction} = 1 \quad (\text{first-order reaction}) \]

### Final Answer
The order of the reaction is **first order**.

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