For the reaction,$NO_{2(g)} + CO_{(g)} \longrightarrow NO_{(g)} + CO_{2(g)}$,the rate of reaction is proportional to the square of $[NO_2]$ and independent of $[CO]$. What is the rate law equation?
- A$r = k \frac{[NO_2]^{1/2}}{[CO]}$
- B$r = k [NO_2]^2 [CO]^0$
- C$r = \frac{1}{2} k [NO_2] [CO]$
- D$r = k [NO_2]^2 [CO]$
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Similar Questions
For a reaction $X + Y \to Z$,the rate is given by $\text{rate} \propto [X]$. What are $(i)$ the molecularity and $(ii)$ the order of the reaction?
Medium
View Solution$A$ reaction $A + B \rightarrow C$ is second order in $A$ and zeroth order in $B$. How is the rate affected when the concentration of $A$ is doubled and $B$ is halved?
The following are the rate constants of two different reactions. Determine the overall order of reaction for each case:
$(a)$ $6.66 \times 10^{-3} \, s^{-1}$
$(b)$ $4.5 \times 10^{-2} \, mol^{-1} \, L \, s^{-1}$
$(a)$ $6.66 \times 10^{-3} \, s^{-1}$
$(b)$ $4.5 \times 10^{-2} \, mol^{-1} \, L \, s^{-1}$
Medium
View SolutionFor the decomposition of a compound $AB$ at $600 \ K$,the following data were obtained:
$[AB] \ (mol \ dm^{-3})$ Rate of decomposition of $AB \ (mol \ dm^{-3} \ s^{-1})$ $0.20$ $2.75 \times 10^{-8}$ $0.40$ $11.0 \times 10^{-8}$ $0.60$ $24.75 \times 10^{-8}$
The order for the decomposition of $AB$ is:
| $[AB] \ (mol \ dm^{-3})$ | Rate of decomposition of $AB \ (mol \ dm^{-3} \ s^{-1})$ |
| $0.20$ | $2.75 \times 10^{-8}$ |
| $0.40$ | $11.0 \times 10^{-8}$ |
| $0.60$ | $24.75 \times 10^{-8}$ |
The order for the decomposition of $AB$ is:
DifficultKCET 2009
View Solution$t_{1/2} =$ constant confirms a first-order reaction. If $a^2 t_{1/2} =$ constant,it confirms that the order of the reaction is ($a =$ initial concentration of reactant).
Medium
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