For the reaction A+B arrow C ; starting with different initial concentration of A and B , initial rate of reaction were determined graphically in four experiments. 3.2× 10- 3 S.NO. [.A].0/M (Initial conc.) [.B].0/M (Initial conc.) rate/M sec-1 1 1.6× 10-3 5× 10-2 10-3 2 3.2× 10-3 5× 10-2 4× 10-3 3 1.6× 10-3 10-1 2× 10-3 4 3.2× 10- 3 10- 1 8× 10- 3 Rate law for reaction from above data is

Question

For the reaction A+B arrow C ; starting with different initial concentration of A and B , initial rate of reaction were determined graphically in four experiments. 3.2× 10- 3 S.NO. [.A].0/M (Initial conc.) [.B].0/M (Initial conc.) rate/M sec-1 1 1.6× 10-3 5× 10-2 10-3 2 3.2× 10-3 5× 10-2 4× 10-3 3 1.6× 10-3 10-1 2× 10-3 4 3.2× 10- 3 10- 1 8× 10- 3 Rate law for reaction from above data is (A) r=k[.A].2[.B].2 (B) r=k[.A].2[.B]. (C) r=k[.A].[.B].2 (D) r=k[.A].[.B].

Tardigrade Admin · Accepted Answer

R=K[A]m[B]n 10-3=K(1.6 × 10-3)m(5 × 10-2)n 3.2 × 10-3=K(3.2 × 10-3)m(5 × 10-2)n 1.6 × 10-3=K(10-1)m(2 × 10-3)n 3.2 × 10-3=K(10-1)m(8 × 10-3)n Solving R=K[A]2[B]

S.NO.	$[A]_{0} / M$ (Initial conc.)	$[B]_{0} / M$ (Initial conc.)	$r a t e / M sec^{- 1}$
$1$	$1.6 \times 10 - 3$	$5 \times 10 - 2$	$10 - 3$
$2$	$3.2 \times 10 - 3$	$5 \times 10 - 2$	$4 \times 10 - 3$
$3$	$1.6 \times 10 - 3$	$10 - 1$	$2 \times 10 - 3$
$4$	$3.2 \times 1 0^{- 3}$	$1 0^{- 1}$	$8 \times 1 0^{- 3}$

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

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

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

$r = k [A] [B]$

r=k[A]2[B]2

r=k[A]2[B]

r=k[A][B]2

r=k[A][B]

R=K[A]m[B]n 10−3=K(1.6×10−3)m(5×10−2)n 3.2×10−3=K(3.2×10−3)m(5×10−2)n 1.6×10−3=K(10−1)m(2×10−3)n 3.2×10−3=K(10−1)m(8×10−3)n Solving R=K[A]2[B]

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

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

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

$r = k [A] [B]$