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If concentration are measured in mole/litre and time in minutes , the unit for the rate constant of a $3^{r d}$ order reaction are

$m o l l i t^{- 1} m i n^{- 1}$

$l i t^{2} m o l^{- 2} m i n^{- 1}$

$l i t m o l^{- 1} m i n^{- 1}$

$m i n^{- 1}$

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Explanation

(B) K= ${[c o n c .]}^{1 - n} m i n^{- 1}$

For $3^{r d}$ order reaction= ${[m o l e / l i t r e]}^{1 - 3} m i n^{- 1}$

= $l i t^{2}, m o l e^{- 2} m i n^{- 1}$

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Following reaction was carried out at 300 K.

$2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$

How is the rate of formation of $S O_{3}$ related to the rate of disapperance of $O_{2}$ ?

$- \frac{∆ [O_{2}]}{∆ t} = + \frac{1}{2} \frac{∆ [S O_{3}]}{∆ t}$

$- \frac{∆ [O_{2}]}{∆ t} = \frac{∆ [S O_{3}]}{∆ t}$

$- \frac{∆ [O_{2}]}{∆ t} = - \frac{1}{2} \frac{∆ [S O_{3}]}{∆ t}$

None of these

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Explanation

(A) Rate of reaction = $- \frac{∆ [O_{2}]}{∆ t} = + \frac{1}{2} \frac{∆ [S O_{3}]}{∆ t}$

Therefore, rate of disappearance of $O_{2}$ is related to rate of formation of $S O_{3}$ as $- \frac{∆ [O_{2}]}{∆ t} = + \frac{1}{2} \frac{∆ [S O_{3}]}{∆ t}$

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Which of the following rate laws has an over all order of 0.5 for the reaction A+B+C $\to$ product-

$R = k [A] . [B] . [C]$

$R = k [A]^{0.5} [B]^{0.5} [C]^{0.5}$

$R = k [A]^{1.5} [B]^{- 1} [C]^{0}$

$R = k [A] [B]^{0} [C]^{0.5}$

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Explanation

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For the system $A_{2} (g) + B_{2} (g) 2 A B (g), ∆ H = - 80 k J$ If the activation energy for the forward step is 100 kcal/mol.What is the ratio of temperature at which the forward and backward reaction shows the same % change of rate constant per degree rise of temperature ? (1 cal= 4.2 J)

$\sqrt{0.72}$

$\sqrt{0.84}$

$\sqrt{0.42}$

$\sqrt{1}$

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Explanation

(B) $E_{a f -} E_{a b} = ∆ H \Rightarrow E_{a b} = E_{a f -} ∆ H = 420 + 80 = 500 k J / m o l N o w, \frac{d (I n k_{f})}{d T} = \frac{E_{a f}}{R T_{f}^{2}} & \frac{d (I n k_{f})}{d T} = \frac{E_{a b}}{R T_{b}^{2}} ∴ \frac{E_{a f}}{T_{f}^{2}} = \frac{E_{a b}}{T_{b}^{2}} \Rightarrow \frac{t_{f}}{t_{b}} = \sqrt{\frac{E_{a f}}{E_{a b}}} = \sqrt{\frac{420}{500}} = \sqrt{0.84}$

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For a reaction of the type 2A+B $\to$ 2C, the rate of the reaction is given by $k {[A]}^{2} [B]$ . When the volume of the reaction vessel is reduced to 1/4 th of the original volume, the rate of reaction changes by a factor of

0.25

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Explanation

(C) Where volume is reduced to 0.25 times the original volume, concentration of each component increases 4 times rate=k ${[4 A]}^{2} [4 B] = k x 64 {[A]}^{2} [B]$

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What is order with respect to A, B, C respectively

[A] [B] [C] rate (M/sec.)

0.2 0.1 0.02 $0.08 \times 10^{- 3}$

0.1 0.2 0.02 $2.01 \times 10^{- 3}$

0.1 1.8 0.18 $6.03 \times 10^{- 3}$

0.2 0.1 0.18 $6.464 \times 10^{- 2}$

$- 1, 1, 3 / 2$

$- 1, 1, 1 / 2$

$1, 3 / 2, - 1$

$1, - 1, 3 / 2$

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Explanation

(D) If rate= $k {[A]}^{x} {[B]}^{y} {[C]}^{z}$

From first two given data

$8.08 \times 10^{- 3} k {[0.2]}^{x} {[0.1]}^{y} {[0.2]}^{z} . . . . . (1) 2.01 \times 10^{- 3} k {[0.1]}^{x} {[0.2]}^{y} {[0.2]}^{z} . . . . . (2)$

Divide $(1) \div (2) w e g e t, 4 = 2^{x} {(1 / 2)}^{y}$

Similarly, from second and third data ${(9)}^{y} {(9)}^{z} = 3 2 y + 2 z = 1$

From first and fourth data $4^{z} = 8 = 2^{3} 2 z = 3, S o z = 3 / 2, y = - 1, x = 1$

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In a reaction, the rate= $k [A] {[B]}^{2 / 3}$ the order of reaction is-

5/3

Zero

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Explanation

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The elementary reaction A+B $\to$ products has $k = 2 \times 10^{- 5} M^{- 1} S^{- 1}$ at a temperature of $27 °$ C. Several experimental runs are carried out using stoichiometric proporation. The reaction has a temperature coefficent value of 2.0. At what temperature should the reaction be carried out if inspite of halving the concentrations, the rate of reaction is desired to be 50% higher than a previous run.

(Given $\frac{l n 6}{l n 2} = 2.585)$

$47 ° C$

$53 ° C$

$57 ° C$

$37 ° C$

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Explanation

(B). $r_{2} = k_{2} {[A]}_{2}^{1} {[B]}_{2}^{1} f o r a c e r t a i n r u n r_{1} = k_{1} {[A]}_{1}^{1} {[B]}_{1}^{1} f o r a p r e v i o u s r u n d i v i d i n g w e g e t, \frac{r_{2}}{r_{1}} = \frac{k_{2}}{k_{1}} \frac{{[A]}_{2}}{{[A]}_{1}} \frac{{[B]}_{2}}{{[B]}_{1}}$

Substituting the given information

$1.5 = 2 (\frac{t_{2} - 27}{10}) \times \frac{1}{2} \times \frac{1}{2} \Rightarrow 6 = 2^{\frac{t_{2} - 27}{10}} \Rightarrow \frac{t_{2} - 27}{10} l n 2 = l n 6 \Rightarrow \frac{t_{2} - 27}{10} = \frac{l n 6}{l n 2} \Rightarrow \frac{t_{2} - 27}{10} = 2.585 \Rightarrow t_{2} = 52.85 ° C ~ 53^{o} C$

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When the temperature of a reaction increases from $27^{0} C$ to $37^{0} C$ , the rate increases by 2.5 times, the activation energy in the temperature range is

53.6 KJ

12.61 KJ

7.08 KJ

70.8 KJ

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Explanation

(D) $\log 2.5 = \frac{E_{a}}{2.303 \times R} \times \{\frac{10}{300 \times 310}\} E_{a} = 70.77 \times 10^{3} J = 70.8 K J$

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For the reaction $R - X + O H \to R O H +$ $X^{-}$ The rate is given of

Rate= $5.0 \times 10^{- 5} [R - X] [O H^{-}] + 0.20 \times 10^{- 5} [R - X]$ what percentage of R-X Reaction by $S N^{2}$ mechanism when $[O H^{-}] = 1.0 \times 10^{- 2} M$

96.1%

3.9%

80%

20%

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Explanation

$R a t e (S N^{2}) = 5.0 \times 10^{5} \times 10^{- 2} [R - X] = 5.0 \times 10^{- 7} [R - X] R a t e (S N^{1}) = 0.20 \times 10^{- 5} [R - X] % o f S N^{2} = \frac{5 \times 10^{- 7} [R - X] \times 100}{5 \times 10^{- 7} [R - X] + 0.20 \times 10^{- 5} [R - X]} = 20$

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