Using the data given below the order and rate constant for the reaction CH3CHOg- CH4g-COg would be-Experiment no-initial conc mol-linitial rate mol-l10-10-0220-20-0830-30-1840-40-32

The correct option is A 2,[k=2.0 M/s] r=k[CH_3CHO]^n→(1) 4r=k[2CH_3CHO]^n→(2) From 1 and 2 eqn 4=2^n ⇒ #160; n= 2 Thus, it is second order ...

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Standard XII

Chemistry

Law of Mass Action

Using the dat...

Question

Using the data given below the order and rate constant for the reaction $C H_{3} C H O (g) \to C H_{4} (g) + C O (g)$ would be:
Experiment no. initial conc (mol/l) initial rate (mol/l)
1 0.1 0.02
2 0.2 0.08
3 0.3 0.18
4 0.4 0.32

2, [k = 2.0 M / s]

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0, [k = 2.0 M / s]

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2, [k = 1.5 M / s]

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1, [

k = 1.5 s^{- 1}

]

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Solution

The correct option is A $2, [k = 2.0 M / s]$
$r = k [C H_{3} C H O]^{n} \to (1)$

$4 r = k [2 C H_{3} C H O]^{n} \to (2)$

From 1 and 2 eqn
$4 = 2^{n} \Rightarrow n = 2$

Thus, it is second-order reaction.

rate $= k [A]^{2}$

k $= \frac{0.02}{(0.1)^{2}} =$ 2.0 l/mol sec.

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Similar questions

Q. Match the rate law given in column I with the dimensions of rate constant given in column II and mark the appropriate choice.

Column I	Column II
(A) $R a t e = k [N H_{3}]^{0}$	(i) $m o l L^{- 1} s^{- 1}$
(B) $R a t e = k [H_{2} O_{2}] [I^{-}]$	(ii) $L m o l^{- 1} s^{- 1}$
(C) $R a t e = k [C H_{3} C H O]^{3 / 2}$	(iii) $s^{- 1}$
(D) $R a t e = k [C_{2} H_{5} C l]$	(iv) $L^{1 / 2} m o l^{- 1 / 2} s^{- 1}$

Q. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i)

2 N O (g) \to N_{2} O (g); R a t e = k [N O]^{2}

(ii)

H_{2} O_{2} (a q) + 3 I^{-} (a q) + 2 H^{+} \to 2 H 2 O (I) + 3 I; R a t e = k [H_{2} O_{2}] [I^{-}]

(iii)

C H_{3} C H O (g) \to C H_{4} (g) + C O (g); R a t e = k [C H_{3} C H O]^{3 / 2}

(iv)

C_{2} H_{5} C l (g) \to C_{2} H_{4} (g) + H C l (g); R a t e = k [C_{2} H_{5} C l]

Q. Identify the order of the reaction from the following rate constants respectively
a.

k = 7.06 \times 10^{- 3} {mol}^{- 1} {L s}^{- 1}

k = 5.60 \times 10^{- 5} s^{- 1}

k = 1.25 \times 10^{- 2} {mol L}^{- 1} s^{- 1}

k = 1.25 \times 10^{- 2} {mol}^{- 2} L^{2} s^{- 1}

Q. Identify the order of the reaction from the following rate constants respectively
a.

k = 7.06 \times 10^{- 3} {mol}^{- 1} {L s}^{- 1}

k = 5.60 \times 10^{- 5} s^{- 1}

k = 1.25 \times 10^{- 2} {mol L}^{- 1} s^{- 1}

k = 1.25 \times 10^{- 2} {mol}^{- 2} L^{2} s^{- 1}

Under the same reaction conditions, the initial concentration of 1.386 mol $d m^{- 3}$ of a substance becomes half in 40 s and 20 s through first order and zero order kinetics, respectively.

The ratio $(k_{1} / k_{0}$ of the rate constants for first order $(k_{1})$ and zero order $(k_{0})$ of the reaction is

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Experiment no.	initial conc (mol/l)	initial rate (mol/l)
1	0.1	0.02
2	0.2	0.08
3	0.3	0.18
4	0.4	0.32

Using the data given below the order and rate constant for the reaction CH3CHO(g)→CH4(g)+CO(g) would be:Experiment no.initial conc (mol/l)initial rate (mol/l)10.10.0220.20.0830.30.1840.40.32

The correct option is A 2,[k=2.0 M/s]r=k[CH3CHO]n→(1)4r=k[2CH3CHO]n→(2)From 1 and 2 eqn4=2n ⇒n=2Thus, it is second-order reaction. rate =k[A]2k=0.02(0.1)2= 2.0 l/mol sec.

Using the data given below the order and rate constant for the reaction $C H_{3} C H O (g) \to C H_{4} (g) + C O (g)$ would be:
Experiment no. initial conc (mol/l) initial rate (mol/l)
1 0.1 0.02
2 0.2 0.08
3 0.3 0.18
4 0.4 0.32

The correct option is A $2, [k = 2.0 M / s]$
$r = k [C H_{3} C H O]^{n} \to (1)$

$4 r = k [2 C H_{3} C H O]^{n} \to (2)$

From 1 and 2 eqn
$4 = 2^{n} \Rightarrow n = 2$

Thus, it is second-order reaction.

rate $= k [A]^{2}$

k $= \frac{0.02}{(0.1)^{2}} =$ 2.0 l/mol sec.