Difficult
For the following rate law,determine the unit of the rate constant: Rate $= -\frac{d[R]}{dt} = k[A]^{\frac{1}{2}}[B]^{2}$
The total order of reaction $n = \frac{1}{2} + 2 = \frac{5}{2} = 2.5$.
Rate $= k[A]^{\frac{1}{2}}[B]^{2} = k[Concentration]^{\frac{5}{2}}$.
$\therefore k = \frac{\text{Rate}}{[Concentration]^{\frac{5}{2}}}$.
$\therefore$ Unit of $k = \frac{\text{mol } L^{-1} s^{-1}}{(\text{mol } L^{-1})^{\frac{5}{2}}} = (\text{mol } L^{-1})^{1 - 2.5} s^{-1} = (\text{mol } L^{-1})^{-1.5} s^{-1}$.
$= \text{mol}^{-1.5} L^{1.5} s^{-1}$ or $L^{1.5} \text{mol}^{-1.5} s^{-1}$.
Rate $= k[A]^{\frac{1}{2}}[B]^{2} = k[Concentration]^{\frac{5}{2}}$.
$\therefore k = \frac{\text{Rate}}{[Concentration]^{\frac{5}{2}}}$.
$\therefore$ Unit of $k = \frac{\text{mol } L^{-1} s^{-1}}{(\text{mol } L^{-1})^{\frac{5}{2}}} = (\text{mol } L^{-1})^{1 - 2.5} s^{-1} = (\text{mol } L^{-1})^{-1.5} s^{-1}$.
$= \text{mol}^{-1.5} L^{1.5} s^{-1}$ or $L^{1.5} \text{mol}^{-1.5} s^{-1}$.
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The following are the rate constants of two different reactions. What is the overall order of reaction for each?
$(a)$ $2.418 \times 10^{-5} \ hr^{-1}$
$(b)$ $7.1 \times 10^{-4} \ atm \ s^{-1}$
$(a)$ $2.418 \times 10^{-5} \ hr^{-1}$
$(b)$ $7.1 \times 10^{-4} \ atm \ s^{-1}$
Medium
View SolutionThe following results have been obtained during the kinetic studies of the reaction:
$2 A + B \rightarrow C + D$
Experiment $[A] / mol \, L^{-1}$ $[B] / mol \, L^{-1}$ Initial rate of formation of $D / mol \, L^{-1} \, min^{-1}$ $I$ $0.1$ $0.1$ $6.0 \times 10^{-3}$ $II$ $0.3$ $0.2$ $7.2 \times 10^{-2}$ $III$ $0.3$ $0.4$ $2.88 \times 10^{-1}$ $IV$ $0.4$ $0.1$ $2.40 \times 10^{-2}$
Determine the rate law and the rate constant for the reaction.
$2 A + B \rightarrow C + D$
| Experiment | $[A] / mol \, L^{-1}$ | $[B] / mol \, L^{-1}$ | Initial rate of formation of $D / mol \, L^{-1} \, min^{-1}$ |
|---|---|---|---|
| $I$ | $0.1$ | $0.1$ | $6.0 \times 10^{-3}$ |
| $II$ | $0.3$ | $0.2$ | $7.2 \times 10^{-2}$ |
| $III$ | $0.3$ | $0.4$ | $2.88 \times 10^{-1}$ |
| $IV$ | $0.4$ | $0.1$ | $2.40 \times 10^{-2}$ |
Determine the rate law and the rate constant for the reaction.
Difficult
View SolutionFor the reaction $Cl_{2(aq)} + H_2S_{(aq)} \to S_{(s)} + 2H^+_{(aq)} + 2Cl^-_{(aq)}$, the rate law is given by $\text{Rate} = K[Cl_2][H_2S]$. Which of the following mechanisms is consistent with this rate law?
$(A)$ $Cl_2 + H_2S \to H^+ + Cl^- + Cl^+ + HS^-$ (slow); $Cl^+ + HS^- \to H^+ + Cl^- + S$ (fast)
$(B)$ $H_2S \rightleftharpoons H^+ + HS^-$ (fast equilibrium); $Cl_2 + HS^- \to 2Cl^- + H^+ + S$ (slow)
$(A)$ $Cl_2 + H_2S \to H^+ + Cl^- + Cl^+ + HS^-$ (slow); $Cl^+ + HS^- \to H^+ + Cl^- + S$ (fast)
$(B)$ $H_2S \rightleftharpoons H^+ + HS^-$ (fast equilibrium); $Cl_2 + HS^- \to 2Cl^- + H^+ + S$ (slow)
Medium
View SolutionFor a certain reaction $A_{(g)} \to B_{(g)}$,the half-life for different initial pressures of $A$ is given below:
$P_{A_0} \text{ (atm)}$ $0.1$ $0.025$ $t_{1/2} \text{ (sec)}$ $100$ $50$
The correct statement about the order of reaction is:
| $P_{A_0} \text{ (atm)}$ | $0.1$ | $0.025$ |
| $t_{1/2} \text{ (sec)}$ | $100$ | $50$ |
The correct statement about the order of reaction is:
Medium
View SolutionThe rate of a reaction is given by $r = K[x][y] / [OH^-]$. If the concentration of $[OH^-]$ is increased,the order of the reaction will be ........
Easy
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