Medium
When the temperature increases from $300 \, K$ to $310 \, K$,the rate of reaction increases by $2.3$ times. If the rate constant at $300 \, K$ is $x$,then the rate constant at $310 \, K$ will be equal to .......
- A$2x$
- B$x$
- C$2.3x$
- D$3x^2$
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Similar Questions
For the reaction $NH_4^+ + NO_2^- \to N_2 + 2H_2O$,the experimental data is given below. Determine the rate law for the reaction.
$1.$ $[NH_4^+] = 0.24 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 7.2 \times 10^{-6} \, M/s$ $2.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 3.6 \times 10^{-6} \, M/s$ $3.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.15 \, M, \text{Rate} = 5.4 \times 10^{-6} \, M/s$
| $1.$ $[NH_4^+] = 0.24 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 7.2 \times 10^{-6} \, M/s$ |
| $2.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 3.6 \times 10^{-6} \, M/s$ |
| $3.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.15 \, M, \text{Rate} = 5.4 \times 10^{-6} \, M/s$ |
Medium
View SolutionUnits of rate constant of first and zero order reactions in terms of molarity $M$ unit are respectively
$A$ chemical reaction takes place in two steps as follows:
$(i) \ NO_2Cl_{(g)} \xrightarrow{K_1} NO_{2(g)} + Cl_{(g)}$
$(ii) \ NO_2Cl_{(g)} + Cl_{(g)} \xrightarrow{K_2} NO_{2(g)} + Cl_{2(g)}$
Identify the reaction intermediate.
$(i) \ NO_2Cl_{(g)} \xrightarrow{K_1} NO_{2(g)} + Cl_{(g)}$
$(ii) \ NO_2Cl_{(g)} + Cl_{(g)} \xrightarrow{K_2} NO_{2(g)} + Cl_{2(g)}$
Identify the reaction intermediate.
The reaction between $X$ and $Y$ is first order with respect to $X$ and zero order with respect to $Y$.
$Experiment$ $[X] / (mol \ L^{-1})$ $[Y] / (mol \ L^{-1})$ $\text{Initial rate} / (mol \ L^{-1} \ min^{-1})$ $I$ $0.1$ $0.1$ $2 \times 10^{-3}$ $II$ $0.2$ $0.2$ $4 \times 10^{-3}$ $III$ $0.4$ $0.4$ $M \times 10^{-3}$ $IV$ $0.1$ $0.2$ $2 \times 10^{-3}$
Examine the data of the table and calculate the ratio of the numerical value of $M$ to $0.2$.
| $Experiment$ | $[X] / (mol \ L^{-1})$ | $[Y] / (mol \ L^{-1})$ | $\text{Initial rate} / (mol \ L^{-1} \ min^{-1})$ |
| $I$ | $0.1$ | $0.1$ | $2 \times 10^{-3}$ |
| $II$ | $0.2$ | $0.2$ | $4 \times 10^{-3}$ |
| $III$ | $0.4$ | $0.4$ | $M \times 10^{-3}$ |
| $IV$ | $0.1$ | $0.2$ | $2 \times 10^{-3}$ |
Examine the data of the table and calculate the ratio of the numerical value of $M$ to $0.2$.
For the reaction $NO_2 + CO \rightarrow CO_2 + NO$,the rate law is given as $\text{Rate} = K [NO_2]^2$. What is the number of $CO$ molecules participating in the slow step?
Easy
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