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)

    tripled

  • Option 3)

    increased by a factor of 4

  • Option 4)

    doubled

 

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

=\frac{dx}{dt}=k(a-x)^{n}

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

t_\frac{1}{2}=\frac{1}{(a^{n-1})(k^{n-1})}[2^{n-1}-1]

 

 

 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)

tripled

This option is incorrect

Option 3)

increased by a factor of 4

This option is correct

Option 4)

doubled

This option is incorrect

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