Solution

Graph (d) represents graph between t_½ and initial concentration for 3rd order reaction :(a) Zero order reaction(b) 1st order reaction(c) 2nd order reaction

In the reaction ,2A + B ⟶ A₂B if the concentration of A is doubled and that of Bis halved,then the rate of the reaction will:

increase 2 times
increase 4 times
decrease 2 times
remain the same

Solution

The rate of reaction is doubled for every 10°C rise in temperature. The increase in reaction rate as a result of temperature rise from 10°C to 100°C is :

Solution

Increase in temperature = 100°C – 10°C = 90°C
∴ n= 9
Increase in reaction rate= 2ⁿ = 2⁹ = 512 times

For reaction aA ⟶ xP , when [A] = 2.2 mM, the rate was found to be 2.4 mMs^-1. On reducing concentration of A to half, the rate changes to 0.6 mMs^-1. The order of reaction with respect to A is :

Solution

When the concentration of reactant is reduced to half its initial value, the rate is reduced by \(\frac{2.4}{0.6}\) = 4 times It means, rate ∝ [ reactant]²
So, order of reaction = 2

For the reaction system : 2NO(g) + O₂(g) ⟶ 2NO₂(g) volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to O2 and second order with respect to NO, the rate of reaction will

diminish to one-eighth of its initial value
increase to eight times of its initial value
increase to four times of its initial value
diminish to one-fourth of its initial value

Solution

r = k[O₂][NO]². When the volume is reduced to 1/2, theconc. will double
∴ New rate = k[2O₂][2NO]² = 8 k [O₂][NO]²
The new rate increases to eight times of its initial.

Solution

Solution

In the given options \(-\frac{d[C]]}{3.dt}\) will not represent the reaction rate. It should not have –ve sign as it is a product. \(\frac{1}{3}\cdot \frac{d[C]]}{dt}\) shows the rate of formation of product C.

Solution

order of reaction is 1.5
molecularity of the reaction is 2
by increasing the concentration of Br₂ four times the rate of reaction is doubled
all the above are correct

Solution

The rate constant of first order reaction is 3 × 10^-6 per second.The initial concentration is 0.10 M. The initial rate is:

3× 10^-7 moles/litre/sec
3 × 10^-8 moles/litre/sec
3× 10^-5 moles/litre/sec
3 × 10^-8 moles/litre/sec

Solution

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