At equilibrium `X+Y hArr 3Z, 1` mol of `X, 2` mol of Y and `4` mol of Z are contained in a `3-L` vessel. What will be the value of the reaction coefficient Q,

Options

A. `10`
B. `15`
C. `10.67`
D. None of these

Explanation

To find the reaction quotient \( Q \) for the reaction \( X + Y \rightleftharpoons 3Z \) at equilibrium, we can follow these steps:

### Step 1: Write the expression for the reaction quotient \( Q \)
The reaction quotient \( Q \) is given by the formula:

\[ Q = \frac{[Z]^3}{[X][Y]} \]

where \([Z]\), \([X]\), and \([Y]\) are the molar concentrations of \( Z \), \( X \), and \( Y \) respectively.

### Step 2: Calculate the concentrations of the reactants and products
We are given the number of moles of each substance and the volume of the vessel. The concentrations can be calculated using the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

- For \( X \):
\[ [X] = \frac{1 \, \text{mol}}{3 \, \text{L}} = \frac{1}{3} \, \text{M} \]

- For \( Y \):
\[ [Y] = \frac{2 \, \text{mol}}{3 \, \text{L}} = \frac{2}{3} \, \text{M} \]

- For \( Z \):
\[ [Z] = \frac{4 \, \text{mol}}{3 \, \text{L}} = \frac{4}{3} \, \text{M} \]

### Step 3: Substitute the concentrations into the expression for \( Q \)
Now we can substitute the calculated concentrations into the expression for \( Q \):

\[ Q = \frac{\left(\frac{4}{3}\right)^3}{\left(\frac{1}{3}\right) \left(\frac{2}{3}\right)} \]

### Step 4: Simplify the expression
Calculating the numerator:

\[ \left(\frac{4}{3}\right)^3 = \frac{64}{27} \]

Calculating the denominator:

\[ \left(\frac{1}{3}\right) \left(\frac{2}{3}\right) = \frac{2}{9} \]

Now substituting back into the expression for \( Q \):

\[ Q = \frac{\frac{64}{27}}{\frac{2}{9}} = \frac{64}{27} \times \frac{9}{2} = \frac{64 \times 9}{27 \times 2} = \frac{576}{54} = \frac{32}{3} \approx 10.67 \]

### Final Answer
Thus, the value of the reaction quotient \( Q \) is approximately **10.67**.

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