The rate low for a 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
B. first
C. second
D. third

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 does not change significantly during the reaction. This means that the concentration of C can be considered approximately constant.

### Step 3: Simplify the rate law
When C is in large excess, we can treat its concentration as a constant. Let’s denote this constant concentration of C as \( [C] = C_0 \). Therefore, we can rewrite the rate law as:
\[ -\frac{d[D]}{dt} = k[C]^2[D] \approx k(C_0)^2[D] \]
Here, \( k' = k(C_0)^2 \) becomes a new effective rate constant.

### Step 4: Identify the new rate law
Now, the rate law simplifies to:
\[ -\frac{d[D]}{dt} = k'[D] \]
where \( k' = k(C_0)^2 \).

### Step 5: Determine the order of the reaction
In this new rate law, the concentration of D is raised to the power of 1. Therefore, the order of the reaction with respect to D is 1.

### Conclusion
Since the order of the reaction is determined by the concentration of D, and C is treated as a constant, the overall order of the reaction is:
\[ \text{Order of reaction} = 1 \]