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The rate of a reaction increases 4-fold when when concentration of reactant is incresed 16 times. If the rate of reaction is $4 \times 10^{- 6} m o l e L^{- 1} S^{- 1} m o l e L^{- 1}$ when concentration of the reactant is $4 \times 10^{- 4}$ , the rate constant of the reaction will be

$2 \times 10^{- 4} m o l e^{1 / 2} L^{- 1 / 2} S^{- 1}$

$1 \times 10^{- 2} S^{- 1}$

$2 \times 10^{- 4} m o l e^{- 1 / 2} L^{- 1 / 2} S^{- 1}$

$25 m o l e^{- 1} L m i n^{- 1}$

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Explanation

(A) $R a t e \propto \sqrt{C o n c e n t t r a t i o n}, R a t e = k \sqrt{C o n c e n t t r a t i o n} k = \frac{r a t e}{{(c o n c e n)}^{1 / 2}} = \frac{4 \times 10^{- 6}}{{(4 \times 10^{- 4})}^{1 / 2}} = \frac{4 \times 10^{- 6}}{2 \times 10^{- 2}} = 2 \times 10^{- 2} m o l e^{1 / 2} L^{- 1 / 2} S^{- 1}$

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If ‘a’ is the initial concentration of a substance which reacts according to zero order kinetic and k is rate constant, the time for the reaction to go to completion is-

a/k

2/ka

k/a

Infinite

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Explanation

(A) For order reaction ${[A]}_{t} = {[A]}_{0} - k t$

For completion of reaction

${[A]}_{t} = 0 a n d t = \frac{{[A]}_{0}}{k} & A_{0} = a; t = \frac{a}{k}$

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The following data are obtained from the decomposition of a gaseous compound

Initial pressure, atm 1.6 0.8 0.4

Time for 50% reaction, min 80 113 160

The order of the reaction is

1.0

1.5

2.0

0.5

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Explanation

(B) $\frac{{(t_{1 / 2})}_{1}}{(t_{1 / 2})} = \frac{80}{113} = {(\frac{p_{2}}{p_{1}})}^{n - 1}$

where $p_{2}, p_{1}$ are the initial pressures

$\frac{80}{113} = {(\frac{0.8}{1.6})}^{n - 1}$

Taking logarithms.

log 0.7=(n-1) log 0.5

solving, n=1.5

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For a given reaction the concentration of the reactant plotted against time gave a straight line with negative slope.

The order of the reaction is-

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Explanation

(D) Rate law is $[A_{0}] - [A] = K t o r [A] = [A_{0}] - K t$

Hence, graph is straight line with negative slope.

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The plot of log k versus $\frac{1}{T}$ is linear with a slope of

$\frac{E_{a}}{R}$

$\frac{- E_{a}}{R}$

$\frac{E_{a}}{2.303 R}$

$\frac{- E_{a}}{2.303 R}$

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Explanation

(D) This is on the basis of Arrhenius equation which is

given us log K= log A $- \frac{E_{a}}{2.303 R T}$

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If a reaction A + B $\to$ C is exothermic to the extent of 30 kJ/mol and the forward reaction has an activation energy 70 kJ/mol, the activation energy for the reverse reaction is

30 kJ/mol

40kJ/mol

70 kJ/mol

100 kJ/mol

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Explanation

For an exothermic reaction, the activation energy for the reverse reaction is higher than the activation energy for the forward reaction by an amount equal to the enthalpy change. Since the forward reaction has an activation energy of 70 kJ/mol and the reaction is exothermic with Î”H = -30 kJ/mol, the activation energy for the reverse reaction will be 70 + 30 = 100 kJ/mol.

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A first order reaction takes 40 min for 30% decomposition.Calculate $t_{1 / 2}$

77.7 min.

27.2 min.

55.3 min.

67.3 min.

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Explanation

(A). Extent of reaction = 30%
Time taken = 40 min
For a first order reaction,

$t = \frac{2.303}{k} \log \frac{a}{a - x} = \frac{2.303}{k} \log \frac{1}{1 - (x / a)} 40 m i n \frac{2.303}{k} \log (\frac{1}{1 - 0.3}) o r k = \frac{2.303}{k} \log \frac{1}{0.7} = 8.92 \times 10^{- 3} m i n^{- 1}$

we know, for a first order reaction, $t_{1 / 2} = \frac{2.303 \log 2}{k}$

then, $t_{1 / 2} = \frac{2.303 \log 2}{8.92 \times 10^{- 3} m i n^{- 1}} = 77.7 m i n .$

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For a certain reaction involving a single reactant, it is found that $C_{0} \sqrt{T}$ is constant where $C_{0}$ is the initial concentration of the reactant and T is the half-life. What is the order of the reaction ?

Zero

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Explanation

(D) $∴ T \propto {(C_{0})}^{1 - n}$

or $T = c o n s \tan t \times {(C_{0})}^{1 - n}$

or $T \times {(C_{0})}^{n - 1} = c o n s \tan t$

or $\sqrt{T \times {(C_{0})}^{n - 1}} = c o n s \tan t$

or $\sqrt{T} \times {(C_{0})}^{\frac{n - 1}{2}} = c o n s \tan t B u t f r o m q u e s t i o n \frac{n - 1}{2} = 1 ∴ n = 3$

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The high temperature ( $\approx$ 1200K) decomposition of $C H_{3} C O O H$ (g) occurs as follows as per simultaneous 1st order reactions.

$C H_{3} C O O H \overset{k_{1}}{\to} C H_{4} + C O_{2}$

$C H_{3} C O O H \overset{k_{2}}{\to} C H_{2} C O + H_{2} O$

What would be the % of $C H_{4}$ by mole in the product mixture (excluding $C H_{3} C O O H$ ) ?

$\frac{50 k_{1}}{(k_{1} + k_{2})}$

$\frac{100 k_{1}}{(k_{1} + k_{2})}$

$\frac{200 k_{1}}{(k_{1} + k_{2})}$

it depends on time

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Explanation

(A) $\frac{n_{C H_{4}} + n_{C O_{2}}}{n_{C H_{2} C O} + n_{H_{2} O}} = \frac{k_{1}}{k_{2}}$

$\Rightarrow \frac{n_{C H_{4}} + n_{C O_{2}}}{n_{C H_{4}} + n_{C O_{2}} + n_{C H_{2} C O} + n_{H_{2} O}} = \frac{k_{1}}{k_{1} + k_{2}} \Rightarrow \frac{2_{n_{C H_{4}}}}{n_{t o t a l}} = \frac{k_{1}}{(k_{1} + k_{2})} \Rightarrow \frac{n_{C H_{4}}}{n_{t o t a l}} = \frac{k_{1}}{2 (k_{1} + k_{2})} \Rightarrow \frac{n_{C H_{4}}}{n_{t o t a l}} \times 100 = \frac{50 k_{1}}{(k_{1} + k_{2})}$

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The rate constant , the activation energy and the Arrhenius parameter of a chemical reaction at 25°C are $3.0 \times 10^{- 4} s^{- 1}$ $, 104.4 k J m o l^{- 4}$ and $6.0 \times 10^{14}$ s–1 respectively the value of the rate constant as T $\to \infty$ is :

$2.0 \times 10^{18} s^{- 1}$

$6.0 \times 10^{- 14} s^{- 1}$

$\infty$

$3.6 \times 10^{30} s^{- 1}$

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Explanation

(B) $K = A e^{- E a / R T} W h e n T \to \infty K \to A A = 6 \times 10^{14} s^{- 1}$

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