Initial concentrations of the reactants and the corresponding half-lives for the reaction P+Q arrow R are given below. The rate law for the reaction is Entry [P_0] (mol dm^ -3 ×10^ -6 ) [ Q_0] ( mol dm^ -3 × 10^ -6 ) t_

Correct answer: A. dR / dt=k [P] No Solution

Olympiad workbook

Olympiad workbook · NSEC · Chemical Kinetics

Initial concentrations of the reactants and the corresponding...

Initial concentrations of the reactants and the corresponding half-lives for the reaction

\mathrm{P}+\mathrm{Q} \rightarrow \mathrm{R}

are given below. The rate law for the reaction is

Entry	\left[P_0\right] \left(mol dm^{-3} \times10^{-6}\right) $</th> <th> \left[$ Q_0\right] \left( $mol dm^{-3}$ \times $10^{-6}\right)$	$t_{1/2}(~s)$
1	500	10	30
2	500	20	60
3	10	500	60
4	20	500	60

A $\mathrm{dR} /$ \mathrm{dt}=\mathrm{k} $[\mathrm{P}]$
B $\mathrm{dR} /$ \mathrm{dt}=\mathrm{k} $[\mathrm{P}] /[\mathrm{Q}]$
C $\mathrm{dR} /$ \mathrm{dt}=\mathrm{k} $[\mathrm{Q}]$
D $-\mathrm{d}[\mathrm{P}] /$ \mathrm{dt}=\mathrm{k} $[\mathrm{P}] /[\mathrm{Q}]$

A. $\mathrm{dR} /$ \mathrm{dt}=\mathrm{k} $[\mathrm{P}]$

No Solution