If the half life of a first order reaction is 6.93 minutes then the time required for completion of $99 \%$ of the reaction will be $\_\_\_\_$ minutes.
(Given $: \log 2=0.3010$ )
$$ \text { For a first order reaction } \mathrm{A} \rightarrow \mathrm{~B} $$
$$ \begin{array}{|l|l|} \hline \mathrm{t} / \min & {[\mathrm{A}] / \mathrm{M}} \\ \hline 0 & 0.6500 \\ \hline x & 0.0650 \\ \hline 20 & 0.00065 \\ \hline \end{array} $$
$x=$ $\_\_\_\_$ min. (Nearest integer)
Consider the following gas phase reaction being carried out in a closed vessel at $25 ^\circ$C.
$$2A(g) \longrightarrow 4B(g) + C(g)$$
| time (min) | total pressure of the system (mm Hg) |
|---|---|
| 30 | 300 |
| ∞ | 600 |
The pressure of C(g) at 30 minutes time interval would be ________ mm Hg. (nearest integer)
For reaction A → P, rate constant $k = 1.5 \times 10^3 \ \mathrm{s}^{-1}$ at $27^{\circ}\mathrm{C}$
If activation energy for the above reaction is $60\ \mathrm{kJ}\ \mathrm{mol}^{-1}$, then the temperature (in $^{\circ}\mathrm{C}$) at which rate constant, $k = 4.5 \times 10^3\ \mathrm{s}^{-1}$ is ______. (Nearest integer)
Given : $\log 2 = 0.30$, $\log 3 = 0.48$, $R = 8.3\ \mathrm{J}\ \mathrm{K}^{-1}\ \mathrm{mol}^{-1}$, $\ln 10 = 2.3$
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