In the chemical reaction, `3A + B to 2C + 3D`, the rate of appearance of C is reported as `1 mol L^(-1)s^(-1)` What is the rate of disappearance of A?

Answer

To solve the problem, we need to relate the rate of disappearance of reactant A to the rate of appearance of product C based on the stoichiometry of the given chemical reaction.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The reaction is given as:
\[ 3A + B \rightarrow 2C + 3D \]

2. **Identify the rates of reaction**:
- The rate of disappearance of A can be denoted as \(-\frac{d[A]}{dt}\).
- The rate of appearance of C can be denoted as \(\frac{d[C]}{dt}\).

3. **Use stoichiometric coefficients to relate the rates**:
According to the stoichiometry of the reaction:
- For every 3 moles of A that disappear, 2 moles of C appear.
- This gives the relationship:
\[ -\frac{1}{3} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[C]}{dt} \]

4. **Rearrange the equation to find the rate of disappearance of A**:
From the above relationship, we can express the rate of disappearance of A in terms of the rate of appearance of C:
\[ \frac{d[A]}{dt} = -\frac{3}{2} \frac{d[C]}{dt} \]

5. **Substitute the given rate of appearance of C**:
We are given that the rate of appearance of C is:
\[ \frac{d[C]}{dt} = 1 \, \text{mol L}^{-1} \text{s}^{-1} \]
Now substitute this value into the equation:
\[ \frac{d[A]}{dt} = -\frac{3}{2} \times 1 = -1.5 \, \text{mol L}^{-1} \text{s}^{-1} \]

6. **Interpret the result**:
The negative sign indicates that A is disappearing. Therefore, the rate of disappearance of A is:
\[ \frac{d[A]}{dt} = 1.5 \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answer:
The rate of disappearance of A is \(1.5 \, \text{mol L}^{-1} \text{s}^{-1}\).