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A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will be

  • Option 1)

    remain unchanged

  • Option 2)


  • Option 3)

    increased by a factor of 4

  • Option 4)



Answers (1)


As we learnt in

nth order reaction -

The rates of the reaction is proportional to nth power of reactant

- wherein

Differential rate law


Integrated rate laws,

=\frac{1}{x-1}=[(a-x)^{1-n} -a^{(1-n)}]=kt

a= initial, concentration of reactant at t=0 sec

x= concentration of product formed at t= tsec




 rate_{1} = K\left [ CO \right ]^{2} \ \\rate_{2} = K\left [ 2CO \right ]^{2}= 4K\left [ CO \right ]^{2} \\rate_{2} = 4\times rate_{1}

Option 1)

remain unchanged

This option is incorrect

Option 2)


This option is incorrect

Option 3)

increased by a factor of 4

This option is correct

Option 4)


This option is incorrect

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