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Byju's Answer

Standard XII

Physics

Time Period Independent of Amplitude

In a reaction...

Question

In a reaction between $A$ and $B$ , initial rate of reaction was measured for different initial concentration $A$ and $B$ given as:

$[A]$ in mol $L^{- 1}$ $[B]$ in mol $L^{- 1}$ Initial rate in mole $L^{- 1} s^{- 1}$
0.2 0.3 $5.07 \times 10^{- 5}$
0.2 0.1 $5.07 \times 10^{- 5}$
0.4 0.05 $1.47 \times 10^{- 4}$

Order of reaction with respect to $A$ and $B$ is:

1, 0

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1, 1

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1.5, 0

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1.5, 1

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Solution

The correct option is C 1.5, 0
$r a t e = k [A]^{m} [B]^{n}$
$5.07 \times 10^{- 5} = k [0.2]^{m} [0.3]^{n}$ ----- 1
$5.07 \times 10^{- 5} = k [0.2]^{m} [0.1]^{n}$ ----- 2
$1.47 \times 10^{- 4} = k [0.4]^{m} [0.05]^{n}$ ----- 3
On solving 1 and 2 equation we get, $n = 0$ .

On solving 1 and 3 equation we get, $m = 1.5$ .
So, $m = 1.5; n = 0$

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$B / m o l L^{- 1}$	$0.30$	$0.10$	$0.05$
$r_{0} / m o l L^{- 1} s^{- 1}$	$5.07 \times 10^{- 5}$	$5.07 \times 10^{- 5}$	$1.43 \times 10^{- 4}$

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A/ mol L⁻¹	0.20	0.20	0.40
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$\begin{matrix} A / m o l L^{- 1} & 0.20 & 0.20 & 0.40 B / m o l L^{- 1} & 0.30 & 0.10 & 0.05 x_{0} / m o l L^{- 1} s^{- 1} & 5.07 \times 10^{- 5} & 5.07 \times 10^{- 5} & 1.43 \times 10^{- 4} \end{matrix}$
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$1.$	$0.2 M$	$0.3 M$	$5 \times 10^{- 5}$
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$[A]$ in mol $L^{- 1}$	$[B]$ in mol $L^{- 1}$	Initial rate in mole $L^{- 1} s^{- 1}$
0.2	0.3	$5.07 \times 10^{- 5}$
0.2	0.1	$5.07 \times 10^{- 5}$
0.4	0.05	$1.47 \times 10^{- 4}$

In a reaction between A and B, initial rate of reaction was measured for different initial concentration A and B given as:[A] in mol L−1[B] in mol L−1Initial rate in mole L−1s−10.20.35.07×10−50.20.15.07×10−50.40.051.47×10−4Order of reaction with respect to A and B is:

The correct option is C 1.5, 0rate=k[A]m[B]n5.07×10−5=k[0.2]m[0.3]n ----- 15.07×10−5=k[0.2]m[0.1]n ----- 21.47×10−4=k[0.4]m[0.05]n ----- 3On solving 1 and 2 equation we get, n=0.On solving 1 and 3 equation we get, m=1.5.So, m=1.5;n=0