Question

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

Answer 1

Hint: The rate of a reaction depends on the concentration of the reactants and the products that participate in the reaction process. The disappearance of the reactant and the increase in the concentration of the product formed have a direct relation with the rate of reaction. Rate of reaction can be calculated by knowing the change in the concentration of the reactants or product in unit time.

Complete step by step answer:
Just like the speed of a vehicle is defined by the distance covered by it in a specific amount of time, the rate of reaction can also be viewed as the change in concentration in time.
Rate of reaction is defined as the change in reaction of reactants or products in unit time. The rate of the reaction can be expressed as the decrease in the concentration of the involved reactants or the rate of increase in the product concentration in the reaction.
Considering the above reaction $A + B \to Products$ we can see that the concentration of A and B will be decreased while the product concentration is bound to be increased. The rate of the expression can be expressed in the terms of either of the two.

For the calculation of the reaction rate we need to divide the moles of the substances that are produced or consumed in the reaction and then dividing by the time of the reaction in seconds
By the above procedure we get,
$A + 3B \to P\,\Delta H = - 2xkj/mol\,of\,A$
$\& M \to 2Q + R\,,\Delta H = + xjk/mol\,of\,M$
Now from the rate law of kinetics,
The disappearance of M $ = $ half the rate of formation of Q
$\dfrac{{ - dM}}{{dt}} = \dfrac{1}{2}\dfrac{{dQ}}{{dt}}$
Since the temperature of the reaction is not changing hence the enthalpies of the reaction is the same so
Rate of disappearance of M should be half the rate of disappearance of A
So, $\dfrac{{ - dA}}{{dt}} = \dfrac{{ - 1}}{3}\dfrac{{dB}}{{dt}}\,\,\,{{since }}\dfrac{{dB}}{{dt}} = Y$
$\dfrac{{ - dM}}{{dt}} = \dfrac{1}{2}\dfrac{{dA}}{{dt}}$
Putting it in the above reaction we get,
$\dfrac{1}{3}\dfrac{{dB}}{{dt}} = \dfrac{1}{2}\dfrac{{dQ}}{{dt}}$
$\dfrac{{dQ}}{{dt}} = \dfrac{2}{3}\dfrac{{dB}}{{dt}} = \dfrac{2}{3}Y$

So, the correct answer is Option A.

Note: The rate of the reaction can be represented in terms of the decrease in the concentration of the reactant as $ - \dfrac{{\Delta [A]}}{{\Delta t}} = - \dfrac{{\Delta [B]}}{{\Delta t}}$ , here the negative sign indicate the decrease in the concentration.