Question

A reaction, 3X(g) $\rightarrow$ 2Y(g) + Z(g) takes place in a closed vessel. What is the rate of formation of Y (in mol L$^{-1}$ s$^{-1}$) if the rate of disappearance of X is 7.2 $\times$ 10$^{-3}$ mol L$^{-1}$ s$^{-1}$?

• A. 3.6 $\times$ 10$^{-3}$
• B. 4.8 $\times$ 10$^{-3}$
• C. 2.4 $\times$ 10$^{-3}$
• D. 1.2 $\times$ 10$^{-3}$

Hint:

Remember to use the stoichiometric coefficients from the balanced chemical equation when relating the rates of disappearance of reactants and the rates of formation of products to the overall rate of the reaction.

Answer 1

The Correct Option is B

Step 1: Write the rate expression for the given reaction.
For the reaction 3X(g) $\rightarrow$ 2Y(g) + Z(g), the rate of reaction can be expressed in terms of the rate of disappearance of the reactant and the rate of formation of the products:
$$ \text{Rate} = -\frac{1}{3} \frac{d[X]}{dt} = +\frac{1}{2} \frac{d[Y]}{dt} = +\frac{d[Z]}{dt} $$
where $\frac{d[X]}{dt}$ is the rate of change of concentration of X, $\frac{d[Y]}{dt}$ is the rate of change of concentration of Y, and $\frac{d[Z]}{dt}$ is the rate of change of concentration of Z. The negative sign indicates disappearance of the reactant, and the positive sign indicates formation of the products.
Step 2: Use the given rate of disappearance of X.
We are given that the rate of disappearance of X is $-\frac{d[X]}{dt} = 7.2 \times 10^{-3}$ mol L$^{-1}$ s$^{-1}$.
Step 3: Relate the rate of disappearance of X to the rate of formation of Y.
From the rate expression, we have:
$$ \frac{1}{2} \frac{d[Y]}{dt} = -\frac{1}{3} \frac{d[X]}{dt} $$ We want to find the rate of formation of Y, which is $\frac{d[Y]}{dt}$. Rearranging the equation:
$$ \frac{d[Y]}{dt} = -\frac{2}{3} \frac{d[X]}{dt} $$ Substitute the given value of $-\frac{d[X]}{dt}$:
$$ \frac{d[Y]}{dt} = \frac{2}{3} \times (7.2 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}) $$ $$ \frac{d[Y]}{dt} = \frac{14.4 \times 10^{-3}}{3} \text{ mol L}^{-1} \text{ s}^{-1} $$ $$ \frac{d[Y]}{dt} = 4.8 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1} $$ Final Answer: \[ \boxed{4.8 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}} \]