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2 months 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)

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Payal Gupta

Contributor-Level 10

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

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

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

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

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

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2 months ago

Chemical Kinetics Class 12th

${[F e {(C N)}_{6}]}^{4 -}$

${[F e {(C N)}_{6}]}^{3 -}$

${[T i {(C N)}_{6}]}^{3 -}$

${[N i {(C N)}_{4}]}^{2 -}$

${[C o {(C N)}_{6}]}^{3 -}$

Among the given complexes, number of paramagnetic complexes in__________.

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alok kumar singh

Contributor-Level 10

CN is a strong field ligand, so pairing occurs.

1. ${[F e {(C N)}_{6}]}^{4 -}$

So; it is diamagnetic

2. ${[F e {(C N)}_{6}]}^{3 -}$

So; it is paramagnetic.

3. ${[T i {(C N)}_{6}]}^{3 -}$

Ti³⁺ = 4s⁰3d¹

It is paramagnetic

4. ${[N i {(C N)}_{4}]}^{2 -}$

It is diamagnetic

5. ${[C o {(C n)}_{6}]}^{3 -}$

Hence, ${[F e {(C N)}_{6}]}^{3 -} a n d {[T i {(C N)}_{6}]}^{3 -}$ ${_{}}^{} {_{}}^{}$ are paramagnetic.

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2 months ago

Chemical Kinetics Class 12th

Spin only magnetic moment of ${[M n B r_{6}]}^{4 -}$ is________ B.M (round off to the closest integer)

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Vishal Baghel

Contributor-Level 10

$M n^{2 +} \overset{}{\to} t_{2 g^{111}} {e_{g}}^{11}$

(Number of unpaired electron = 5)

$μ_{S} = \sqrt[]{35} = 5.91 ≅ 6$

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3 months ago

Chemical Kinetics Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

40% of HI undergoes decomposition to H₂ and I₂ at 300K. $Δ G^{Θ}$ for this decomposition reaction at one atmosphere pressure is_____________J mol^-1 (nearest integer)

(Use R = 8.31 JK^-1 mol^-1; log 2= 0.3010, In 10 = 2.3, log 3 = 0.477)

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Vishal Baghel

Contributor-Level 10

$H l ? \frac{1}{2} H_{2} + \frac{1}{2} l_{2}$

t = eq1 - $\frac{α}{2}$ $\frac{α}{2} [α = 0.4]$

$K p = \frac{{(P_{H_{2}})}^{\frac{1}{2}} * {(P_{l_{2}})}^{\frac{1}{2}}}{P_{H l}} = \frac{{(0.2)}^{\frac{1}{2}} * {(0.2)}^{\frac{1}{2}}}{0.6}$

= $- 8.31 * 300 * 2.3 * l o g (\frac{1}{3}) = 2735 J / m o l$

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3 months ago

Chemical Kinetics Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

The activation energy of one of the reactions is a biochemical process is 532611 J mol^-1 when the temperature falls from 310 K to 300 K, the change in rate constant observed is k₃₀₀ = x * 10^-3 k₃₁₀. The value of x is___________.

[Given: In 10 = 2.3 R = 8.3 JK^-1 mol^-1]

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Vishal Baghel

Contributor-Level 10

$l n (\frac{K_{2}}{K_{1}}) = \frac{E a}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l n (\frac{K_{2}}{K_{1}}) = \frac{532611}{8.3} * (\frac{10}{310 * 300})$

Where, K₂ is at 310 K and K₁ at 300K

$l n (\frac{K_{2}}{K_{1}}) = 6.9 = 3 * l n 10$

$l n (\frac{K_{2}}{K_{1}}) = l n 1 0^{3}$

K₂ = K₁ * 10³

K₁ = K₂ * 10^-3

$∴$ x = 1

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3 months ago

Chemical Kinetics Chemistry NCERT Exemplar Solutions Class 12th Chapter One

For a reaction A--> 2B + C the half lives are 100s and 50s when the concentration of reactant A is 0.5 and 1.0 mol L^-1 respectively. The order of the reaction is _____________. (Nearest (Integer)

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Vishal Baghel

Contributor-Level 10

$t_{1 / 2} α {[R e a c t a n t]}_{t = 0}^{(1 - n)},$ where n = order of reaction.

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3 months ago

Chemical Kinetics Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

For a reaction, given below is the graph of ink vs $\frac{1}{T} .$ The activation energy for the reaction is equal to_____________ cal mol^-1 (Nearest integer)

(Given: R = 2 cal K^-1 mol^-1)

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Vishal Baghel

Contributor-Level 10

$k = A e^{- E a / R T}$

$l n k = l n A - \frac{E a}{R T}$

Comparing with y = mx + c

Slope = $+ \frac{E a}{R} = + \frac{20}{5}$

$∴ E a = \frac{20}{5} R = 4 R = 4 * 2 c a l$

= 8 Cal

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3 months ago

Chemical Kinetics Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

If $75 %$ of a first order reaction was completed in 90 minutes, $60 %$ of the same reaction would be completed in approximately (in minutes)

(Take: $l o g 2 = 0.30; l o g 2.5 = 0.40$ )(Kinetics)

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alok kumar singh

Contributor-Level 10

First order reaction

$K = \frac{2.303}{t} l o g ? \frac{a_{0}}{a_{0} - x}$

$K = \frac{2.303}{90} l o g ? \frac{a_{0}}{0.25 a_{0}}$

$= 0.0154$

$\begin{array}{r} t = 60 % = \frac{2.303}{K} l o g ? \frac{a_{0}}{a_{0}} \dots . (2) \\ = \frac{2.303}{0.0154} * (1 - 0.602) = 59.51 m i n s \approx 60 \end{array}$

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3 months ago

Chemical Kinetics Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

For a first order reaction A→B, the rate constant, k = 5.5 * 10^-¹⁴s^-¹. The time required for 67% completion of reaction is x * 10^-¹ times the half life of reaction. The value of x is ________ (Nearest integer)

(Given : log 3 = 0.4771)

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Payal Gupta

Contributor-Level 10

Rate constant, k = 5.5 * 10^-14 s^-1.

$t = \frac{2.303}{k} l o g \frac{{[R]}_{0}}{[R]}$

= $\frac{2.303}{k} l o g 3 - (i)$

$t_{50 %} = \frac{2.303}{k} l o g 2 - (i i)$

From (i) & (ii)

$\frac{t_{67 %}}{t_{50 %}} = \frac{l o g 3}{l o g 2}$

= 1.58 t_50%

So; t_67%is 15.8 * 10^-1 times half life.

X = 16 (the nearest integer)

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3 months ago

Chemical Kinetics Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

The reaction between X and Y is first order respect to X and zero order with respect to Y.

Examine the data of table and calculate ratio of numerical values of M and L (Nearest integer)

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Vishal Baghel

Contributor-Level 10

According to condition

Rate $\propto {[X]}^{1} {[Y]}^{0}$

By using experimental data I and II

$\frac{4 * 1 0^{- 3}}{2 * 1 0^{- 3}} = \frac{L}{0.1} ∴ L = 0.2$

And by using experimental data I and II

$\frac{M * 1 0^{- 3}}{2 * 1 0^{- 3}} = \frac{0.4}{0.1}$

$∴$ M = 8

Ratio of $(\frac{M}{L}) = \frac{8}{0.2} = 40$

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