Chemical Kinetics

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New answer posted•a year ago

Chemical Kinetics Class 12th

If the activation energy of a reaction is 80.9 kJ mol^-1, the fraction of molecules at 700K, having enough energy to react to form products, is e^-x. The value of x is………. (Rounded-off to the nearest integer).

[use R = 8.31 J K^-1 mol^-1]

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Fraction of molecules having enough energy to form product = $e^{- E a / R T}$ $^{}$

Fraction of molecules having enough energy to form product = $e^{- \frac{80.9 \times 1 0^{3}}{8.314 \times 700}}$ $^{\frac{^{}}{}}$

$= e^{- 13.8} ≅ e^{- 14}$

So, x = 14

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New answer posted•a year ago

Chemical Kinetics Class 12th

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half – life of 3.33 h at 25°C. After 9h, the fraction of sucrose remaining is f. The value of log₁₀ $(\frac{1}{f})$  is………………….× 10^-². (Rounded off to the nearest integer)

[Assume: In 10 = 2.303, In2 = 0.693]

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$C_{12} H_{22} O_{11} + H_{2} O \to \underset{G l u c o s e}{C_{6} H_{12} O_{6}} + \underset{F r u c t o s e}{C_{6} H_{12} O_{6}}$

$K = \frac{2.303}{t} l o g \frac{a}{a - x}$

$\frac{k t}{2.303} = l o g \frac{a}{a - x}$

$\frac{l n 2 \times 9}{\frac{10}{3} \times 2.303} = l o g (\frac{1}{f})$

$l o g (\frac{1}{f}) = 81.24 \times 1 0^{- 2} = 81 \times 1 0^{- 2}$

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New answer posted•a year ago

Chemical Kinetics Class 12th

For the reaction aA + bB->cC + dD, the plot of log K vs $\frac{1}{T}$ is given below:

This temperature at which the rate constant of the reaction is 10^-4 s^-1 is----------K.

(Rounded – off to the nearest integer)

[Given : The rate constant of the reaction is 10^-5 s^-1 at 500 K]

13 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$l o g K = l o g A - \frac{E a}{2.303 R T}$

$S l o p e = \frac{E a}{2.303 R} = - 10, 000 K$

$\frac{E a}{2.303 R} = 1 0^{4}$

$l o g (1 0^{- 5}) = l o g A - 1 0^{4} \times \frac{1}{500}$ (at 500 K temperature)

$∴ T = \frac{1 0^{4}}{19} K = \frac{100}{19} \times 1 0^{2} K = 5.2631 \times 1 0^{2} K = 526.31 K \approx 526 K$

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New question posted•a year ago

Chemical Kinetics Class 12th

Benzene on nitration givens nitrobenzene in presence of HNO3 and H2SO4 mixture, where:

2 Views|Posted a year ago

Asked by Vishal Baghel

New answer posted•a year ago

Chemical Kinetics Class 12th

For the reaction A -> B, the rate constant K(in s^-1) is givne by

$l o g_{10} K = 20.35 - \frac{(2.47 \times 1 0^{3})}{T} .$

The energy of activation in KJ mol^-1 is ______. (Nearest integer)

[Given : R = 8.314 J K^-1 mol^-1]

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

log k = 20.35 - $\frac{(2.47 \times 1 0^{3})}{T}$

Comparing with,

$l o g k = l o g A - \frac{E a}{2.303 R T}$

$\frac{E a}{2.303 R} = 2.47 \times 1 0^{3}$

= 47.29 kJ/mole

Ans. = 47

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New answer posted•a year ago

Chemical Kinetics Class 12th

At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340s. When the pressure was 27.8 kPa., the half life was found to be 170s. The order of the reaction is ________. [integer answer]

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

t_1/2 = 340 sec P₀ = 55.5 kPa

t_1/2 = 170 sec P₀ = 27.8 kPa

$? t_{1 / 2} ? {(P_{0})}^{1 ? n}$

$? 1 ? n = 1 n = 0$

$?$ zero order reaction

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New answer posted•a year ago

Chemical Kinetics Class 12th

[A] -> [B]

Reactant Product

If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x × 10^-⁶ s^-¹. The value of x is___________. (Nearest Integer)

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$K = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{70 \times 60} = \frac{6930}{7 \times 6} \times 1 0^{- 6}$

= 165 × 10^-6 s^-1

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New answer posted•a year ago

Chemical Kinetics Class 12th

The half life for the decomposition of gaseous compound A is 240s when the gaseous pressure was 500 Torr initially, When the pressure was 250 Torr, the half life was found to be 4.0 min. The order of the reaction is _________. (Nearest integer)

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{{(t_{1 / 2})}_{l}}{{(t_{1 / 2})}_{I I}} = {(\frac{P_{I}}{P_{I I}})}^{1 - n} \Rightarrow n = 1$

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New answer posted•a year ago

Chemical Kinetics Class 12th

For the given first order reaction

A -> B

The half life of the reaction is 0.3010 min. The ration of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to________________. (Nearest integer)

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

t1/2 = 0.301 min

t = 2 min

$K = \frac{2.303}{t} l o g (\frac{C o}{C t})$

$\frac{0.693}{0.301} = \frac{2.303}{2} l o g (\frac{C o}{C t})$

$\frac{2.303 \times 0.301}{0.301} = \frac{2.303}{2} l o g (\frac{C o}{C t})$

$∴ 2 = l o g (\frac{C o}{C t})$

$\frac{C o}{C t} = 1 0^{2} = 100$

Ans. 100

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New answer posted•a year ago

Chemical Kinetics Class 12th

For a given chemical reaction

$γ_{1} A + γ_{2} B \to γ_{3} C + γ_{4} D$ 

Concentration of C charges form 10 mmol dm^-³ to 20 mmol dm^-³ in 10 second. Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance A. The rate of appearance of D has been experimentally determined to be 9 mmol dm^-³ s^-¹. Therefore the rate of reaction is_____________mmol dm^-³ s^-¹ (Nearest Integer)

19 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$\frac{d [C]}{d t} = (\frac{20 - 10}{10}) = 1 m m o l d m^{- 3}$

$+ \frac{d [D]}{d t} = {- \frac{d (B)}{d t}} \times 1.5 = \frac{3}{2} {- \frac{d [B]}{d t}}$

${- \frac{d [B]}{d t}} = 2 \times {\frac{- d [A]}{d t}}$

$∴ \frac{1}{6} {\frac{- d [B]}{d t}} = \frac{1}{3} {\frac{- d [A]}{d t}} = \frac{1}{9} {\frac{+ d [D]}{d t}} = \frac{d [C]}{d t}$

$∴$ rate of reaction = $\frac{+ d [C]}{d t}$ = 1m.m dm^-3 S^-1

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