The reaction,$X + 2Y + Z \to N$ occurs by the following mechanism:
$(i)$ $X + Y \rightleftharpoons M$ (very rapid equilibrium)
$(ii)$ $M + Z \to P$ (slow)
$(iii)$ $P + Y \to N$ (very fast)
What is the rate law for this reaction?

  • A
    Rate $= k[Z]$
  • B
    Rate $= k[X][Y]^2[Z]$
  • C
    Rate $= [N]$
  • D
    Rate $= k[X][Y][Z]$

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Consider the following reaction: $A \longrightarrow \text{Products}$. This reaction is completed in $100 \ min$. The rate constant of this reaction at $t_1 = 10 \ min$ is $10^{-2} \ min^{-1}$. What is the rate constant (in $min^{-1}$) at $t_2 = 20 \ min$?

In a reaction between $A$ and $B$,the initial rate of reaction $(r_0)$ was measured for different initial concentrations of $A$ and $B$ as given below:
$A / mol \ L^{-1}$ $0.20$ $0.20$ $0.40$
$B / mol \ L^{-1}$ $0.30$ $0.10$ $0.05$
$r_0 / mol \ L^{-1} \ s^{-1}$ $5.07 \times 10^{-5}$ $5.07 \times 10^{-5}$ $1.43 \times 10^{-4}$

What is the order of the reaction with respect to $A$ and $B$?

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For a hypothetical reaction,$A \rightarrow C$. The mechanism is: $A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} B$ (Fast),$A + B \xrightarrow{k_3} C$ (Slow). The rate law for this reaction is:

The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the following table:
Experiment $[A] / mol \, L^{-1}$ $[B] / mol \, L^{-1}$ Initial rate / $mol \, L^{-1} \, min^{-1}$
$I$ $0.1$ $0.1$ $2.0 \times 10^{-2}$
$II$ $-$ $0.2$ $4.0 \times 10^{-2}$
$III$ $0.4$ $0.4$ $-$
$IV$ $-$ $0.2$ $2.0 \times 10^{-2}$

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