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Class 10 Class 12

The decomposition of a compound is found to follow a first-order rate law. If it takes 15 minutes for 20 percent of original material to react.
Calculate :

(i) the specific rate constant
(ii) the time at which 10 percent of the original material remains unreacted
(iii) the time it takes for the next 20 percent of the reactant left to react after the first 15 minutes.

According to the first order rate -law

(i) k = \frac{2.303}{t} \log \frac{a}{a - x} k = \frac{2.303}{t} \log \frac{100}{80} k = 0.154 \log 1.25 k = 0.154 \times 0.0969 k = 1.50 \times 10^{- 1} \min^{- 1}

Time at which 10 percent of the original material remains unreacted is ,

(ii) t = \frac{2.303}{1.50 \times 10^{- 2}} \times \log \frac{100}{10} t = 1.54 \times 10^{2} \min = 154 \min .

Time for the next 20 percent reaction

(iii) t = \frac{2.303}{1.50 \times 10^{- 2}} \times \log \frac{100}{60} t = 1.54 \times 10^{2} [\log 10 - \log 6] = 154 \times [1.0 - 0.7782] = 34.15 \min .

472 Views

The rate of a particular reaction doubles when temperature changes from 27°C to 37°C. Calculate the activation energy of such reaction.

According to the Arrhenius equation

\log \frac{k_{2}}{k_{1}} = \frac{Ea}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}}) \log \frac{1}{2} = \frac{Ea}{2.303 \times 8.314} \times (\frac{310 - 300}{310 \times 300}) Ea = \frac{19.147 \times 93000 \times 0.3010}{10} = \frac{19.147 \times 9300 \times 0.3010}{1000} Ea = 52.99 kJ {mol}^{- 1} .

153 Views

Thermal decomposition of a compound is of first order. If 50% of a sample of the compound is decomposed in 120 minutes, how long will it take for 90% of the compound to decompose.

For half time period

k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{120} = 5.77 \times 10^{- 3} \min^{- 1}

For a first order reaction

k = \frac{2.303}{t} \log \frac{1}{a - x} a = 100, x = 90 ∴ 5.77 \times 10^{- 3} = \frac{2.303}{t} \log \frac{100}{100 - 90} t = \frac{2.303}{5.77 \times 10^{- 3}} \log 10 = 399 \min .

1133 Views

Rate constant k of a reaction varies with temperature according to the equation:

log k constant $- \frac{E_{a}}{2.303} \frac{1}{T}$

where E_a is the energy of activation for the reaction. When a graph is plotted for log k versus 1/T a straight line with a slope - 6670 k is obtained. Calculate energy of activation for this reaction. State the units (R = 8.314 J K^–1 mol^–1)

We have given the slope value ,
Slope = -6670 k

$E_{a} = 2 R = 8.314 J K^{- 1} {mol}^{- 1}$

Thus using the equation

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

$\Rightarrow E_{a} = - slope \times 2.303 \times R = - (- 6670) \times 2.303 \times 8.314 J = + 127710.49 J = + 127.71049 kJ .$

472 Views

A first order reaction takes 69.3 minutes for 50% completion. Set up an equation determining the time needed for 80% completion of the reaction.

Half life reaction can be given by
t_1/2 = 0.693/k
thus using above equation we get

t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{69.3} = \frac{69.3 \times 10^{- 2}}{69.3 \min} = 1 \times 10^{- 2} \min u \sin g f i r s t o r d e r e q u a t i o n, w e g e t K = \frac{2.303}{t} \log \frac{[R_{0}]}{[R]} t = \frac{2.303}{1 \times 10^{- 2}} \log \frac{[R_{0}]}{\frac{20}{180} [R_{0}]} t = \frac{2.303}{1 \times 10^{- 2}} \log 5 \frac{2.303}{10^{- 2}} \times 0.6990 = 1.6097 \times 10^{2} \min = 160.97 \min .

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