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Standard XII

Chemistry

Molecularity of Reaction

The rate of t...

Question

The rate of the reaction, $2 N O + O_{2} ⟶ 2 N O_{2}$ , at $25^{o} C$ is $0.028$ $m o l$ $L^{- 1} s^{- 1}$ . The experimental rate is given by, $r = k {[N O]}^{2} [O_{2}]$ .
If the initial concentrations of the reactants are $O_{2} = 0.040$ $m o l$ $L^{- 1}$ and $N O = 0.01$ $m o l$ $L^{- 1}$ , the rate constant of the reaction is:

7.0 \times 10^{- 2}

L

{m o l}^{- 1} s^{- 1}

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7.0 \times 10^{- 4} L^{2} {m o l}^{- 2} s^{- 1}

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7.0 \times 10^{2} L^{2} {m o l}^{- 2} s^{- 1}

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7.0 \times 10^{3} L^{2} {m o l}^{- 2} s^{- 1}

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Solution

The correct option is D $7.0 \times 10^{3} L^{2} {m o l}^{- 2} s^{- 1}$
$r = k [N O]^{2} [O_{2}]$
Now, using given quantities:
$0.028 = k . (0.04) (0.01) k = 7 \times 10^{- 3} L^{2} {m o l}^{- 2} s^{- 1}$

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Similar questions

Q. The reaction

2 N O + O_{2} \to 2 N O_{2}

, follows the rate law

= k {[N O]}^{2} [O_{2}]

. What is the order of the reaction? If

k = 2.0 \times 10^{- 6} {m o l}^{- 2} L^{2} s^{- 1}

, what is the rate of the reaction when

[N O] = 0.04

m o l

L^{- 1}

and

[O_{2}] = 0.2

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Q. In a reaction,

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}

, the rate of disappearance of

N_{2} O_{5}

6.5 \times 10^{- 3}

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. Compute the rates of formation of

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Q. For the reaction

A \to B

, rate constant

K = 2 \times 10^{- 4} m o l L^{- 1} S^{- 1}

. If initial concentration of

A

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concentration of

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after

20

minute is:

For the reaction, $2 A + B \to A_{2} B$ the rate $= k [A] [B]^{2}$ with $k = 2.0 \times 10^{- 6} m o l^{- 2} L^{- 2} s^{- 1}$ . Calculate the initial rate of reaction when $[A] = 0.1 m o l L^{- 1}, [B] = 0.2 m o l L^{- 1}$ . Calculate the rate of reaction after [A] is reduced to $0.06 m o l L^{- 1}$ .

Q. For the reaction:

2 A + B \to A_{2} B

, the rate

= k [A] [B]^{2}

with

k = 2.0 \times 10^{- 6} m o l^{- 2} L^{2} s^{- 1}

Calculate the initial rate of the reaction when

[A] = 0.1 m o l L^{- 1}, [B] = 0.2 m o l L^{- 1}

. Calculate the rate of reaction after

[A]

is reduced to

0.06 m o l L^{- 1}

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The rate of the reaction, 2NO+O2⟶2NO2, at 25oC is 0.028 mol L−1s−1. The experimental rate is given by, r=k[NO]2[O2].If the initial concentrations of the reactants are O2=0.040 mol L−1 and NO=0.01 mol L−1, the rate constant of the reaction is:

The correct option is D 7.0×103 L2 mol−2s−1r=k[NO]2[O2] Now, using given quantities:0.028=k.(0.04)(0.01)k=7×10−3L2mol−2s−1

The correct option is D $7.0 \times 10^{3} L^{2} {m o l}^{- 2} s^{- 1}$
$r = k [N O]^{2} [O_{2}]$
Now, using given quantities:
$0.028 = k . (0.04) (0.01) k = 7 \times 10^{- 3} L^{2} {m o l}^{- 2} s^{- 1}$