Consider the following reaction which proceeds in a closed vessel.
` 3X to 2Y +Z `
The rate of disappearance of `X,-(Delta [X])/(Delta t)` is found to be 0.075 mol
` L^(-) s^(-)` calculate `(Delta [y])/(Delta t) and ( Delta [Z])/( Delta t)`

Explanation

To solve the problem, we need to analyze the stoichiometry of the given reaction and apply the relationship between the rates of disappearance and appearance of the reactants and products.

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction is given as:
\[ 3X \rightarrow 2Y + Z \]

2. **Identify the Rate of Disappearance of X**:
We are given that the rate of disappearance of X is:
\[ -\frac{\Delta [X]}{\Delta t} = 0.075 \, \text{mol L}^{-1} \text{s}^{-1} \]

3. **Relate the Rate of Disappearance of X to the Rate of Appearance of Y and Z**:
From the stoichiometry of the reaction, we can express the rates of change of concentrations as follows:
\[ -\frac{1}{3} \frac{\Delta [X]}{\Delta t} = \frac{1}{2} \frac{\Delta [Y]}{\Delta t} = \frac{1}{1} \frac{\Delta [Z]}{\Delta t} \]

4. **Calculate the Rate of Appearance of Y**:
Using the relationship for Y:
\[ \frac{\Delta [Y]}{\Delta t} = -\frac{2}{3} \frac{\Delta [X]}{\Delta t} \]
Substituting the value of \(-\frac{\Delta [X]}{\Delta t}\):
\[ \frac{\Delta [Y]}{\Delta t} = -\frac{2}{3} \times (-0.075) = \frac{2 \times 0.075}{3} = \frac{0.15}{3} = 0.05 \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Calculate the Rate of Appearance of Z**:
Using the relationship for Z:
\[ \frac{\Delta [Z]}{\Delta t} = -\frac{1}{3} \frac{\Delta [X]}{\Delta t} \]
Substituting the value of \(-\frac{\Delta [X]}{\Delta t}\):
\[ \frac{\Delta [Z]}{\Delta t} = -\frac{1}{3} \times (-0.075) = \frac{0.075}{3} = 0.025 \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answers:
- The rate of appearance of Y is:
\[ \frac{\Delta [Y]}{\Delta t} = 0.05 \, \text{mol L}^{-1} \text{s}^{-1} \]
- The rate of appearance of Z is:
\[ \frac{\Delta [Z]}{\Delta t} = 0.025 \, \text{mol L}^{-1} \text{s}^{-1} \]