**
**

**Q1. **For the reaction,
2N_2 O_5 (g)→4NO_2 (g)+O_2 (g)
If the concentration of NO_2 increase by 5.2×〖10〗^(-3) M in 100 s then the rate of the reactions
Solution

Rate of reaction =1/4 d(NO_2 )/dt=(5.2×〖10〗^(-3))/(4×100)
=1.3×〖10〗^(-5 ) Ms^(-1)

**Q2.**For a fist order reaction, the concentration changes from 0.8 to 0.4 in 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is
Solution

T_50=15 min
k=2.303log2/T_50 =2.303log2/15
a=0.1M
(a-x)=0.025 M
For first order reaction,
k=2.303log2/T_50 log(a/(a-x))
2.303log2/15=(2×2.303log2)/t log 0.1/0.025
=2.303/t log4
∴2.303log2/15=(2×2.303log2)/t
∴ t=30 min

**Q3. **If the concentration units are reduced by n times, then the value of rate constant of first order will:
Solution

Rate constant is characteristic constant of a reaction and depends only on temperature and catalyst.

**Q4. **Inversion of cane-sugar in dilute acid is a
Solution

c_12 H_22 O_11+H_2 O□(→┴(Dil.acid) ) C_6 H_12 O_6+C_6+H_12 O_6
Excess glucose fructose
when one of the reactant is present in large excess, the second order reaction confirms to the first order and is knows as Pseudo-unimolecular reaction

**Q5.**The unit of rate constant for a zero order reaction
Solution

C

**Q6. **For the reaction; 2N_2 O_5⟶4NO_2+O_2, rate and rate constant are 1.02×〖10〗^(-4) M sec^(-1)and 3.4×〖10〗^(-5) sec^(-1)M respectively, then concentration of N_2 O_5, at that time will be:
Solution

r=K[N_2 O_5]
∴[N_2 O_5 ]=r/K=(1.02×〖10〗^(-4))/(3.4×〖10〗^(-5) )=3M
.

**Q7.**If “a” and “t_(1/2)” are initial concentration of reactant and half-life of a zero order reaction respectively, which of the following is correct ?
Solution

For zero order reaction integrated rate equation is
kt=[A]_0-[A]
If 〖[A]〗_0=a,[A]=a/2,t=t_(1/2)
kt_(1/2)=a-a/2
kt_(1/2)=a/2
t_(1/2)=a/2k
∴ t_(1/2)∝a

**Q8.**In the reaction 2A+B→A_2 B, if the concentration of A is doubled and of B is halved, then the rate of the reaction will
Solution

2A+B→A_2 B
r_1=k[A]^2 [B]
r_2=k[2A]^2 [B/2]
Or r_2=2k[A]^2 [B]
∴r_2=2r_1

**Q9.**The Arrhenius equation expressing the effect of temperature on the rate constant of reaction is:
Solution

This is Arrhenius equation.

**Q10. **The rate constant is given by the equation K=Ae^(-E_a/RT) which factor should register a decrease for the reaction to proceed more rapidly?

Solution

A decrease in E_a will increase rate constant K and thus rate of reaction increases.

**Q1.**For the reaction, 2N_2 O_5 (g)→4NO_2 (g)+O_2 (g) If the concentration of NO_2 increase by 5.2×〖10〗^(-3) M in 100 s then the rate of the reactions

Solution

Rate of reaction =1/4 d(NO_2 )/dt=(5.2×〖10〗^(-3))/(4×100) =1.3×〖10〗^(-5 ) Ms^(-1)

Rate of reaction =1/4 d(NO_2 )/dt=(5.2×〖10〗^(-3))/(4×100) =1.3×〖10〗^(-5 ) Ms^(-1)

**Q2.**For a fist order reaction, the concentration changes from 0.8 to 0.4 in 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is

Solution

T_50=15 min k=2.303log2/T_50 =2.303log2/15 a=0.1M (a-x)=0.025 M For first order reaction, k=2.303log2/T_50 log(a/(a-x)) 2.303log2/15=(2×2.303log2)/t log 0.1/0.025 =2.303/t log4 ∴2.303log2/15=(2×2.303log2)/t ∴ t=30 min

T_50=15 min k=2.303log2/T_50 =2.303log2/15 a=0.1M (a-x)=0.025 M For first order reaction, k=2.303log2/T_50 log(a/(a-x)) 2.303log2/15=(2×2.303log2)/t log 0.1/0.025 =2.303/t log4 ∴2.303log2/15=(2×2.303log2)/t ∴ t=30 min

**Q3.**If the concentration units are reduced by n times, then the value of rate constant of first order will:

Solution

Rate constant is characteristic constant of a reaction and depends only on temperature and catalyst.

Rate constant is characteristic constant of a reaction and depends only on temperature and catalyst.

**Q4.**Inversion of cane-sugar in dilute acid is a

Solution

c_12 H_22 O_11+H_2 O□(→┴(Dil.acid) ) C_6 H_12 O_6+C_6+H_12 O_6 Excess glucose fructose when one of the reactant is present in large excess, the second order reaction confirms to the first order and is knows as Pseudo-unimolecular reaction

c_12 H_22 O_11+H_2 O□(→┴(Dil.acid) ) C_6 H_12 O_6+C_6+H_12 O_6 Excess glucose fructose when one of the reactant is present in large excess, the second order reaction confirms to the first order and is knows as Pseudo-unimolecular reaction

**Q5.**The unit of rate constant for a zero order reaction

Solution

C

C

**Q6.**For the reaction; 2N_2 O_5⟶4NO_2+O_2, rate and rate constant are 1.02×〖10〗^(-4) M sec^(-1)and 3.4×〖10〗^(-5) sec^(-1)M respectively, then concentration of N_2 O_5, at that time will be:

Solution

r=K[N_2 O_5] ∴[N_2 O_5 ]=r/K=(1.02×〖10〗^(-4))/(3.4×〖10〗^(-5) )=3M .

r=K[N_2 O_5] ∴[N_2 O_5 ]=r/K=(1.02×〖10〗^(-4))/(3.4×〖10〗^(-5) )=3M .

**Q7.**If “a” and “t_(1/2)” are initial concentration of reactant and half-life of a zero order reaction respectively, which of the following is correct ?

Solution

For zero order reaction integrated rate equation is kt=[A]_0-[A] If 〖[A]〗_0=a,[A]=a/2,t=t_(1/2) kt_(1/2)=a-a/2 kt_(1/2)=a/2 t_(1/2)=a/2k ∴ t_(1/2)∝a

For zero order reaction integrated rate equation is kt=[A]_0-[A] If 〖[A]〗_0=a,[A]=a/2,t=t_(1/2) kt_(1/2)=a-a/2 kt_(1/2)=a/2 t_(1/2)=a/2k ∴ t_(1/2)∝a

**Q8.**In the reaction 2A+B→A_2 B, if the concentration of A is doubled and of B is halved, then the rate of the reaction will

Solution

2A+B→A_2 B r_1=k[A]^2 [B] r_2=k[2A]^2 [B/2] Or r_2=2k[A]^2 [B] ∴r_2=2r_1

2A+B→A_2 B r_1=k[A]^2 [B] r_2=k[2A]^2 [B/2] Or r_2=2k[A]^2 [B] ∴r_2=2r_1

**Q9.**The Arrhenius equation expressing the effect of temperature on the rate constant of reaction is:

Solution

This is Arrhenius equation.

This is Arrhenius equation.

**Q10.**The rate constant is given by the equation K=Ae^(-E_a/RT) which factor should register a decrease for the reaction to proceed more rapidly?

Solution

A decrease in E_a will increase rate constant K and thus rate of reaction increases.

A decrease in E_a will increase rate constant K and thus rate of reaction increases.