The energies of activation for forward and reverse reactions for A₂ + B₂ ⟶ 2AB are 180 kJ mol^-1 and 200 kJ mol^-1 respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol^-1. The enthalpy change of the reaction (A₂ + B₂ ⟶ 2AB) in the presence of a catalyst will be (in kJ mol^-1)

Solution

Presence of catalyst does not affect enthalpy change of reaction ∆H_R = E_f - E_b = 180– 200 = – 20 kJ/mol

a
b
c
d
b & d both

Solution

In a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become :

256 times
512 times
64 times
128 times

Solution

The unit of rate constant for a zero order reaction is

mol L^-1 s^-1
L mol^-1 s^-1
L² mol^-1 s^-1
s^-1

Solution

Rate = k[A]°
Unit of k = mol L^-1 sec^-1

Which one of the following statements for the order of a reaction is incorrect ?

Order can be determined only experimentally.
Order is not influenced by stoichiometric coefficient of the reactants.
Order of reaction is sum of power to the concentration terms of reactants to express the rate of reaction.
Order of reaction is always whole number

Solution

order of reaction may be zero, whole number or fractional.

The rate of the reaction 2NO + Cl₂ ⟶ 2NOCl is given by the rate equation rate= k [NO]²[Cl₂]The value of the rate constant can be increased by:

increasing the concentration of NO.
increasing the temperature.
increasing the concentration of the Cl₂
doing all of these

Solution

2NO(g) + Cl₂(g) ⇌ 2NOCl(g)Rate = k[NO]² [Cl]The value of rate constant can be increased by increasing the temperature.
∴ Correct choice :(b)

rate = k[A]²[B]
rate = k[A] [B]
rate =k[A]²[B]²
rate =k[A] [B]²

Solution

In case of(II) and(III), keeping concentration of [A]constant, when the concentration of [B]is doubled,the rate quadruples. Hence it is second order with respect to B. In case of I & IV Keeping the concentration of [B] constant.when the concentration of [A] is increased four times,rate also increases four times.Hence, the order with respect to A is one. hence
Rate = k[A][B]²

For an endothermic reaction, energy of activation is E_a and enthalpy of reaction of ΔH(both of these in kJ/mol).Minimum value of E_a will be.

less than ΔH
equal to ΔH
more than ΔH
equal to zero

For the reaction [N₂O₅(g) ⟶ 2NO₂(g) + 1/2 O₂(g)]the value of rate of disappearance of N₂O₅ is given as 6.25× 10^-3 mol L^-1 s^-1. The rate of formation of NO₂ and O₂ is given respectively as :

6.25 × 10^-3 mol L^-1 s^-1 and 6.25 × 10^-3 mol L^-1s^-1
1.25 × 10^-2 mol L^-1 s^-1 and 3.125 × 10^-3 mol L^-1 s^-1
6.25 × 10^-3 mol L^-1 s^-1 and 3.125 × 10^-3 mol L^-1 s^-1
1.25 × 10^-2 mol L^-1 s^-1 and 6.25 × 10^-3 mol L^-1 s^-1

Solution

For the reaction A + B ⟶ products, it is observed that:
(1)On doubling the initial concentration of A only, the rate of reaction is also doubled and
(2)On doubling the initial concentrations of both A and B,there is a change by a factor of 8 in the rate of the reaction.The rate of this reaction is given by:

rate = k[A] [B]²
rate = k[A]² [B]²
rate = k[A] [B]
rate = k[A]² [B]

Solution

When concentration A is doubled, rate is doubled.Hence order with respect to A is one.When concentrations of both A and B are doubled,rate increases by 8 times hence order with respect to Bis 2.
∴ rate= k[A]¹[B]² Total order = 1 + 2 = 3

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