The rate law for the reaction below is given by the expression $k\left[ A \right]\left[ B \right]$
$A + B \to$ Product
If the concentration of $B$ is increased from $0.1$ to $0.3\, mole$, keeping the value of $A$ at $0.1\, mole$, the rate constant will be
$3k$
$9k$
$k/3$
$k$
For the reaction $A + 2B \to C,$ rate is given by $R$ $ = [A]{[B]^2}$ then the order of the reaction is
The rate constant for the reaction $2N_2O_5 \to 4NO_2 + O_2$ is $3.0\times10^{-5}\, sec^{-1}$. If rate is $2.40\times10^{-5}\, M\, sec^{-1}$, then the concentration of $N_2O_5$ (in $M$) is ?
Write unit of rate constant of following reaction :
$1.$ $\frac {5}{2}$ order
$2.$ $n$ order
Select the rate law that corresponds to the data shown for the following reaction $A+ B\to C$
Expt. No. | $(A)$ | $(B)$ | Initial Rate |
$1$ | $0.012$ | $0.035$ | $0.10$ |
$2$ | $0.024$ | $0.070$ | $0.80$ |
$3$ |
$0.024$ |
$0.035$ | $0.10$ |
$4$ | $0.012$ | $0.070$ | $0.80$ |
The reaction $CH _{3} COF + H _{2} O \quad \rightleftharpoons CH _{3} COOH + HF$
Condition $I$ $:$ $\left[ H _{2} O \right]_{0}=1.00 \,M$
$\left[ CH _{3} COF \right]_{0}=0.01 \,M$
Condition $II$ $:$ $\left[ H _{2} O \right]_{0}=0.02 \,M$
$\left[ CH _{3} COF \right]_{0}=0.80 \,M$
Condition - $I$ | Condition - $II$ | ||
Time $min$ |
$\left[ CH _{3} COF \right]$ $M$ |
Time $min$ |
$\left[ H _{2} O \right] \,M$ |
$0$ | $0.01000$ | $0$ | $0.0200$ |
$10$ | $0.00867$ | $10$ | $0.0176$ |
$20$ | $0.00735$ | $20$ | $0.0156$ |
$40$ | $0.00540$ | $40$ | $0.0122$ |
Determine the order of reaction and calculate rate constant.