Understanding the Reaction and Kinetics

For the reaction: $A\left(g\right)\to 2B\left(g\right)$, we analyze the kinetics. Let [A] be the concentration of A at time t, [B] be the concentration of B at time t, and a be the initial concentration of A.

Step 1: Relate [B] to [A]

Since 1 mole of A produces 2 moles of B, the concentration of B at any time is given by: $\left[B\right]=2(a-[A\left]\right)$.

Step 2: Analyze Each Case

(A) Zero Order Kinetics: $\frac{d\left[B\right]}{dt}$ vs $\frac{-d\left[A\right]}{dt}$ curve

For zero order, $\frac{-d\left[A\right]}{dt}=k$ (constant). From [B] = 2(a - [A]), differentiate: $\frac{d\left[B\right]}{dt}=2\frac{-d\left[A\right]}{dt}=2k$. So, $\frac{d\left[B\right]}{dt}$ is constant and directly proportional to $\frac{-d\left[A\right]}{dt}$. The graph is a straight line through origin with slope 1 (since if x-axis is $\frac{-d\left[A\right]}{dt}$ and y-axis is $\frac{d\left[B\right]}{dt}$, then y = 2x, but note the curve in column II: the first image shows a straight line through origin, which matches). So, (A) matches the first image.

(B) First Order Kinetics: [B] vs t curve

For first order, $\left[A\right]=a{e}^{-kt}$. Then [B] = 2(a - [A]) = 2a(1 - e^-kt). As t increases, [B] increases and approaches 2a asymptotically. The curve is exponential rise, similar to the second image in column II.

(C) First Order Kinetics: Half-life vs initial concentration

For first order reaction, half-life $t_{1/2}=\frac{\ln 2}{k}$, which is independent of initial concentration. So, the graph is a horizontal straight line (constant). The third image in column II shows a horizontal line, so (C) matches.

(D) Zero Order Kinetics: Half-life vs initial concentration

For zero order, $t_{1/2}=\frac{a}{2k}$. So, half-life is directly proportional to initial concentration a. The graph is a straight line through origin with positive slope. The fourth image in column II shows a straight line through origin, so (D) matches.

Final Answer:

(A) matches the first curve, (B) matches the second curve, (C) matches the third curve, (D) matches the fourth curve.

Related Topics and Formulae

Zero Order Kinetics

Rate is constant: $\frac{-d\left[A\right]}{dt}=k$. Integrated rate law: $\left[A\right]=[A{]}_0-kt$. Half-life: $t_{1/2}=\frac{[A{]}_0}{2k}$.

First Order Kinetics

Rate is proportional to concentration: $\frac{-d\left[A\right]}{dt}=k\left[A\right]$. Integrated rate law: $\ln \left[A\right]=\ln [A{]}_0-kt$. Half-life: $t_{1/2}=\frac{\ln 2}{k}$.

Stoichiometry in Kinetics

For reaction $aA\to bB$, the rates are related by $\frac{1}{a}\frac{-d\left[A\right]}{dt}=\frac{1}{b}\frac{d\left[B\right]}{dt}$.