## In uranium mineral, the atomic ratio $\frac{N^{U-238}}{N^{Pb-206}}$ is nearly equal to one…

Problem : In uranium mineral, the atomic ratio $\frac{N^{U-238}}{N^{Pb-206}}$ is nearly equal to one. The age (in years) of the mineral is nearly (given that half-life of U238 is 4.5 × 109 years).

(A) 3.0 × 108

(B) 4.5 × 108

(C) 3.0 × 109

(D) 4.5 × 109

Solution: Equal no. of atoms of U238 and Pb206 means half of U has distintegrated

time taken = t1/2

Ans: (D)

## In the nuclear reaction 235U92 → 207Pb82 , the number of α and β particles lost would be

Problem : In the nuclear reaction 235U92207Pb82 , the number of α and β particles lost would be

(A) 8, 4

(B) 6, 2

(C) 7, 4

(D) 4, 3

Solution: 235U92207Pb82 + x 4α2 + y 0β-1

235 = 207 + 4x

⇒ x = 7

92 = 82 + 2x – y

⇒  y = 4

Ans: (C)

## $SO_2Cl_2 \rightleftharpoons SO_2 + Cl_2$ is the first order gas reaction with…

Problem : $SO_2Cl_2 \rightleftharpoons SO_2 + Cl_2$ is the first order gas reaction with K = 2.2 × 10–5 sec–1 at 320°C. The percentage of SO2Cl2 decomposed on heating for 90 minutes is

(A) 1.118

(B) 0.1118

(C) 18.11

(D) 11.30

Solution: $\large K = \frac{2.303}{t} log\frac{a}{a-x}$

$\large log\frac{a}{a-x} = \frac{K t}{2.303}$

$\large log\frac{a}{a-x} = \frac{2.2 \times 10^{-5}\times 60 \times 60}{2.303} = 0.0516$

$\large \frac{a}{a-x} = 1.127$

$\large \frac{a-x}{a} = 0.887$

$\large 1- \frac{x}{a} = 0.887$

$\large \frac{x}{a} = 0.113$

= 11.3 %

Ans: (D)

## The rate of reaction is doubled for every 10° rise in temperature. The increase in reaction rate…

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

(A) 112

(B) 512

(C) 400

(D) 614

Solution: Increase in steps of 10° has been made 9 times. Hence rate of reaction should increase 29 times i.e. 512 times.

(B)

## The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction…

Problem : The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25°C are 3.0 × 10–4 s–1 , 104.4 kJ mol–1 and 6.0 × 1014 s–1 respectively the value of the rate constant as T →∞ is

(A) 2.0 × 1018 s–1

(B) 6.0 × 1014 s–1

(C) ∞

(D) 3.6 × 1030 s–1

Solution: $\large K = A e^{E_a/RT}$

When T → ∞

K → A

A = 6 × 1014 s–1