The reaction 2NO- O 2 - 2N O 2- follows the rate law -k- NO - 2 - O 2 - What is the order of the reaction- If k-2-0- 10 -6 mol -2 L 2 s -1- what is the rate of the reaction when - NO - -0-04 mol L -1 and - O 2 - -0-2 mol L -1-

Order =2+1=3 Rate #160; =2× 10 ^ 6 ×[ 0.04 ] #160; ^ 2 ×[ 0.2 ] #160; #160; #160; #160; #160; =6.4× 10 ^ 10 mol L^ 1 s ^ 1 ...

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Standard XII

Chemistry

Molecularity of Reaction

The reaction ...

Question

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 $[O_{2}] = 0.2$ $m o l$ $L^{- 1}$ ?

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Solution

Order $= 2 + 1 = 3$
Rate $= 2 \times 10^{- 6} \times {[0.04]}^{2} \times [0.2]$
$= 6.4 \times 10^{- 10} m o l L^{- 1} s^{- 1}$

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Q. The rate of the reaction,

2 N O + O_{2} ⟶ 2 N O_{2}

, at

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0.028

m o l

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. The experimental rate is given by,

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

.
If the initial concentrations of the reactants are

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, the rate constant of the reaction 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}

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Q. For the reaction:

2 H_{2} + 2 N O ⇌ 2 H_{2} O + N_{2}

, the following rate data was obtained:

S. No.	$[N O]$ $m o l$ $L^{- 1}$	$[H_{2}]$ $m o l$ $L^{- 1}$	Rate: $m o l$ $L^{- 1}$ ${s e c}^{- 1}$
$1$	$0.40$	$0.40$	$4.6 \times 10^{- 3}$
$2$	$0.80$	$0.40$	$18.4 \times 10^{- 3}$
$3$	$0.40$	$0.80$	$9.2 \times 10^{- 3}$

Calculate the following:
(1) The overall order of reaction.
(2) The rate law.
(3) The value of rate constant

(k)

Q. The mechanism of the reaction

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

is:

N O + N O_{1} N_{2} O_{2} (f a s t); N_{2} O_{2} + O_{2}_{2} 2 N O_{2} (s l o w)

The rate constant of the reaction is:

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The reaction 2NO+O2→2NO2, follows the rate law =k[NO]2[O2]. What is the order of the reaction? If k=2.0×10−6mol−2L2s−1, what is the rate of the reaction when [NO]=0.04 mol L−1 and [O2]=0.2 mol L−1?

Order =2+1=3Rate =2×10−6×[0.04]2×[0.2] =6.4×10−10molL−1s−1

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 $[O_{2}] = 0.2$ $m o l$ $L^{- 1}$ ?

Order $= 2 + 1 = 3$
Rate $= 2 \times 10^{- 6} \times {[0.04]}^{2} \times [0.2]$
$= 6.4 \times 10^{- 10} m o l L^{- 1} s^{- 1}$