What is the rate law expression for the reaction, $A + B \to C$ ?
The following data were collected from the experiments:
Expt.
No. Initial $[A]$
$(m o l L^{- 1})$ Initial $[B]$
$(m o l L^{- 1})$ Rate of
formation of $C$
$1.$ $0.10$ $0.10$ $4.0 \times 10^{- 4}$
$2.$ $0.20$ $0.20$ $3.2 \times 10^{- 3}$
$3.$ $0.10$ $0.20$ $1.6 \times 10^{- 3}$

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Solution

Rate $= K {[A]}^{m} {[B]}^{n}$

${(\frac{0.10}{0.10})}^{m} {(\frac{0.10}{0.100})}^{n} = \frac{4.0 \times 10^{- 4}}{1.6 \times 10^{- 3}}$

${(0.5)}^{n} = 2.5 \times 10^{- 3} n = 2$

${(\frac{0.20}{0.10})}^{m} {(\frac{0.20}{0.20})}^{n} = \frac{3.2 \times 10^{- 3}}{1.6 \times 10^{- 3}}$

$m = 1$

Rate $= K {[A]}^{1} {[B]}^{2}$

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Similar questions

A + 2 B ⇌ C

In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B?

A

B

(r_{0})

A

B

A

B

A + B \to C

2 A + B \to

Experment	[ $A$ ] (in mol $L^{- 1}$ )	[ $B$ ] (in mol $L^{- 1}$ )	Initial Rate of reaction (in mol $L^{- 1} m i n^{- 1}$ )
(I)	$0.10$	$0.20$	$6.93 \times 10^{- 3}$
(II)	$0.10$	$0.25$	$6.93 \times 10^{- 3}$
(III)	$0.20$	$0.30$	$1.386 \times 10^{- 2}$

A

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