Question

The initial rate of zero order reaction of the gaseous reaction $A(g) \longrightarrow 2B(g)$ is $10^{-2} M \ min^{-1}$. If the initial concentration of A is 0.1 M, after 60 s concentration of B would be

• A. 0.09 M
• B. 0.01 M
• C. 0.02 M
• D. 0.002 M

Answer 1

Answer: (C)

0.02 M

Answer 2

For a zero-order reaction, the rate of reaction is constant and equal to the rate constant ($k$). Given the initial rate as $10^{-2} M \ min^{-1}$, $k = 10^{-2} M \ min^{-1}$.

The amount of reactant A consumed in time $t$ is given by $kt$.

Converting $t = 60 \ s$ to minutes, we get $t = 1 \ min$.

Amount of A consumed = $(10^{-2} M \ min^{-1}) \times (1 \ min) = 0.01 M$.

From the reaction stoichiometry, $A(g) \longrightarrow 2B(g)$, 1 mole of A produces 2 moles of B.

Therefore, if 0.01 M of A is consumed, the concentration of B formed will be $2 \times 0.01 M = 0.02 M$.