Question

For a hypothetical reaction,
${\text{A}} + {\text{3B}} \to {\text{P }}\Delta {\text{H}} = - 2{\text{x kJ/mol of A}}$
${\text{M}} \to {\text{2Q}} + {\text{R }}\Delta {\text{H}} = + {\text{x kJ/mol of M}}$
If these reactions are carried out simultaneously in a reactor such that temperature is not changing. If rate of disappearance of B is ${\text{y M se}}{{\text{c}}^{ - 1}}$ then rate of formation (in ${\text{M se}}{{\text{c}}^{ - 1}}$) of Q is:
(A) $\dfrac{2}{3}{\text{y}}$
(B) $\dfrac{3}{2}{\text{y}}$
(C) $\dfrac{4}{3}{\text{y}}$
(D) $\dfrac{3}{4}{\text{y}}$

Answer 1

Hint: Here we have to calculate the rate of the reaction. To calculate the rate of the reaction, divide the number of moles of substance produced or consumed in the reaction by the time required for the reaction to complete in seconds.

Complete step by step answer:
We are given two reaction as follows:
${\text{A}} + {\text{3B}} \to {\text{P}}\Delta {\text{H}} = - 2{\text{x kJ/mol of A}}$ …… Reaction (1)
${\text{M}} \to {\text{2Q}} + {\text{R}}\Delta {\text{H}} = + {\text{x kJ/mol of M}}$ …… Reaction (2)
In reaction (2), M is getting consumed and Q is being produced. The rate of disappearance of M is equal to the half of the rate of Q being produced with a change in sign.
Thus, in the reaction (2), the rate of disappearance of M is as follows:
$-\dfrac{{{\text{dM}}}}{{{\text{dt}}}}=\dfrac{1}{2}\dfrac{{{\text{dQ}}}}{{{\text{dt}}}}$ …… (1)
In reaction (1), A is getting consumed and B is also getting consumed. The rate of disappearance of A is equal to the one third of the rate of disappearance of B. Thus,
$ - \dfrac{{{\text{dA}}}}{{{\text{dt}}}} = - \dfrac{1}{3}\dfrac{{{\text{dB}}}}{{{\text{dt}}}}$ …… (2)
We are given that the temperature is not changing. At constant temperature, the change in enthalpy is equal. Thus,
$\Delta {{\text{H}}_1} = \Delta {{\text{H}}_2}$
Thus, the rate of disappearance of M is equal to the half of the rate of disappearance of A. Thus,
$\dfrac{{{\text{dM}}}}{{{\text{dt}}}} = \dfrac{1}{2}\dfrac{{{\text{dA}}}}{{{\text{dt}}}}$
Thus, equation (1) becomes,
$\dfrac{1}{3}\dfrac{{{\text{dB}}}}{{{\text{dt}}}} = \dfrac{1}{2}\dfrac{{{\text{dQ}}}}{{{\text{dt}}}}$
Thus,
$\dfrac{{{\text{dQ}}}}{{{\text{dt}}}} = \dfrac{2}{3}\dfrac{{{\text{dB}}}}{{{\text{dt}}}}$
We are given that the rate of disappearance of B is ${\text{y M se}}{{\text{c}}^{ - 1}}$. Thus,
$\dfrac{{{\text{dB}}}}{{{\text{dt}}}} = {\text{y}}$
Thus,
$\dfrac{{{\text{dQ}}}}{{{\text{dt}}}} = \dfrac{2}{3}{\text{y}}$
Thus, the rate of formation of Q is $\dfrac{2}{3}{\text{y}}$.

Thus, the correct option is (A) $\dfrac{2}{3}{\text{y}}$.

Note: The rate of the reaction is the speed at which the reactants are being consumed or the products are being formed. The rate of any reaction is expressed in terms of concentration of a product formed in a unit of time or in terms of concentration of a reactant consumed in a unit of time.