The following initial rate data were obtained for the reaction- 2NO g - Br 2 g - 2NOBr g Run - NO - - M - Br 2 - - M Rate - M min -1 1- 1-0 1-0 1-30- 10 -3 2- 2-0 1-0 5-20- 10 -3 3- 4-0 2-0 4-16- 10 -2Determine the reaction rate law and the value of the rate constant-

Rate #160; =K[ NO ] #160; ^ m [ B r _ 2 ] #160; ^ n ( 1 2 #160; ) #160; ^ m ( 1 1 #160; ) #160; ^ n = 1.30× 10 ^ 3 5.20× 10 ^ 3 ...

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Byju's Answer

Standard XII

Chemistry

Arrhenius Equation

The following...

Question

The following initial rate data were obtained for the reaction:
$2 N O (g) + {B r}_{2} (g) \to 2 N O B r (g)$
Run $[N O] / M$ $[{B r}_{2}] / M$ $R a t e / M {m i n}^{- 1}$
$1.$ $1.0$ $1.0$ $1.30 \times 10^{- 3}$
$2.$ $2.0$ $1.0$ $5.20 \times 10^{- 3}$
$3.$ $4.0$ $2.0$ $4.16 \times 10^{- 2}$
Determine the reaction rate law and the value of the rate constant.

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Solution

Rate $= K {[N O]}^{m} {[B r_{2}]}^{n}$

${(\frac{1}{2})}^{m} {(\frac{1}{1})}^{n} = \frac{1.30 \times 10^{- 3}}{5.20 \times 10^{- 3}}$
$m = 2$

${(\frac{1}{4})}^{m} {(\frac{1}{2})}^{n} = \frac{1.30 \times 10^{- 3}}{4.16 \times 10^{- 2}}$
${(\frac{1}{4})}^{2} {(\frac{1}{2})}^{n} = \frac{1.30 \times 10^{- 3}}{4.16 \times 10^{- 2}}$
$n = 1$

Rate $= K {[N O]}^{2} {[{B r}_{2}]}^{1}$
$K = ?$
$1.30 \times 10^{- 3} = K {(1)}^{2} \times 1$
$K = 1.30 \times 10^{- 3} {m o l}^{- 2} L^{2} s^{- 1}$

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0.1	0.1	$1.3 \times 10^{- 6}$
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Run	$[N O] / M$	$[{B r}_{2}] / M$	$R a t e / M {m i n}^{- 1}$
$1.$	$1.0$	$1.0$	$1.30 \times 10^{- 3}$
$2.$	$2.0$	$1.0$	$5.20 \times 10^{- 3}$
$3.$	$4.0$	$2.0$	$4.16 \times 10^{- 2}$

The following initial rate data were obtained for the reaction:2NO(g)+Br2(g)→2NOBr(g)Run[NO]/M[Br2]/MRate/Mmin−11.1.01.01.30×10−32.2.01.05.20×10−33.4.02.04.16×10−2Determine the reaction rate law and the value of the rate constant.

Rate =K[NO]m[Br2]n(12)m(11)n=1.30×10−35.20×10−3m=2(14)m(12)n=1.30×10−34.16×10−2(14)2(12)n=1.30×10−34.16×10−2n=1Rate =K[NO]2[Br2]1K=?1.30×10−3=K(1)2×1K=1.30×10−3mol−2L2s−1