Question

A first order reaction: $\longrightarrow$ products and a second order reaction: $\longrightarrow$ $2R\overset{\quad\quad}{\rightarrow}$Products both have half - time of 20 minutes when they are carried out taking 4 mole L-1 of there respective reactants. Number of mole per litre of A and R remaining unreacted after 60 minutes from the start of the reaction, respectively will be.

• A. 1 and 0.5
• B. 0.5 and negligible
• C. 0.5 and 1
• D. 1 and 0.25

Answer 1

Answer: (C)

0.5 and 1

Answer 2

In case of first order reaction t1/2 will remain constant independent of initial concentration so.

4 mole L $^ { - 1 } \xrightarrow { 20 \min } 2$ mole L$^ { - 1 } \xrightarrow { 20 \mathrm {~min} }$ $1 \mathrm {~mole} \mathrm {~L} ^ { - 1 } \xrightarrow { 20 \mathrm {~min} } 0.5 \mathrm { moleL } ^ { - 1 }$

That is, after 60 minutes 0.5 mole L-1 of A will be left unreacted.

In case of second order reaction t1/2 is inversely proportional to initial concentration of reactant i.e., t1/2 will go on doubling as concentration of reactant will go on getting half. That is, t1/2a will be constant, so.

$\mathbf{4mole}\mathbf{L}^{\mathbf{-}\mathbf{1}}\overset{\mathbf{\quad}\mathbf{20}\mathbf{\min}\mathbf{\rightarrow}^{\mathbf{-}\mathbf{1}}\overset{\mathbf{\quad}\mathbf{20}\mathbf{\min}\mathbf{\rightarrow}^{\mathbf{-}\mathbf{1}}\mathbf{\quad}}{\rightarrow}\mathbf{\quad}}{\rightarrow}$That is, after 60 min, the concentration of R remaining unreacted will be 1 mole L-1