Question

How many times must the concentration of substance $B_2$ in the system $2A_2\left(g\right)+B_2\left(g\right)\to 2A_{2}B\left(g\right)$ be increased for the rate of the forward reaction to remain unchanged when the concentration of substance $A_2$ is lowered to one-fourth of its initial value?

Answer 1

$2A_2\left(g\right)+B_2\left(g\right)⟶2A_{2}B\left(g\right)$

$K_P=\frac{[A_{2}B{]}^2}{\left[A_2{]}^2\right[B_2]}$ or

$K_P=\frac{{P_{A_{2}B}}^2}{{P_{A_2}}^2{P}_{B_2}}⟶\left(1\right)$

when, $A_2^1=\frac{1}{4}{A}_2$,

then, $K_P=\frac{{P_{A_2}B}^2}{P_{A_2}^1{P}_{B_2}^1}=\frac{{P_{A_{2}B}}^2}{{\left(\frac{1}{4}{P}_{A_2}\right)}^2{P}_{B_2}^1}⟶\left(2\right)$

From (1) & (2),

$\frac{1}{P_{A_2}^2{P}_{B_2}}=\frac{1}{\frac{1}{16}{P}_A^2{P}_{B_2}^1}$

$\therefore P_{B_2}^1=16P_{B_2}$

$\therefore B_2$ is to be increased $16$ times.