Question

plz solve this......☺☺​

Answer 1

Answer :-

$(1)\to \boxed{ \sf{ rate = 8 \times {10}^{ - 9} \: {mol}^{} {L}^{-1} \: {s}^{ - 1} }} \: \\ \\ (2)\to \boxed{ \sf{ rate = 38.88 \times {10}^{ - 10} {mol}^{} {L}^{-1} \: {s}^{ - 1} }} \:$

To Find :-

(1) Initial rate of reaction for given data .

(2) Calculate rate of reaction after [A] is reduced to 0.06 mol L^{-1}

Explanation :-

Given that

$\to \sf{k = 2.0 \times {10}^{ - 6} {mol}^{ - 2} {L }^{2} {s}^{ - 1} } \\ \\ (1) \sf{ Initial \: rate \: when \: [A] = \: 0.1 \: mol {L}^{ - 1} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{[B] = 0.2 mol {L}^{-1} \: } \\ \\ \to \sf{rate = k[A]{[B]}^{2} \: } \\ \\ \sf{= (2.0 \times {10}^{ - 6} {mol}^{ - 2} {L }^{2} {s}^{ - 1})(0.1mol {L}^{-1}) {(0.2 mol {L}^{-1})}^{2} } \\ \\ \to \sf{rate = 2.0 \times {10}^{ - 6} \times 0.1 \times 0.04 \: \: {mol}^{} {L}^{-1} \: {s}^{ - 1} } \\ \\ \to \boxed{ \sf{ rate = 8 \times {10}^{ - 9} \: {mol}^{} {L}^{-1} \: {s}^{ - 1} }}$

(2) Rate of reaction when [A] reduced to 0.06 mol L^(-1)

So concentration of [A] = 0.1 - 0.06 = 0.04 mol L^(-1)

Therefore concentration of [B] is

[B] = half of [A] = 0.02

→ so available [B] = 0.2 - 0.02 = 0.18

$\to \sf{}rate \: \sf{= (2.0 \times {10}^{ - 6} {mol}^{ - 2} {L }^{2} {s}^{ - 1})(0.06mol {L}^{-1}) {(0.18mol {L}^{-1})}^{2} } \: \\ \\ \to \sf{ rate = 2.0 \times {10}^{ - 6} \times 0.06 \times 0.0324 \: {mol}^{2} {L}^{-1} \: {s}^{ - 1} } \: \\ \\ \to \sf{rate = 0.003888 \times {10}^{ - 6} \: {mol}^{2} {L}^{-1} \: {s}^{ - 1} } \: \\ \\ \to \boxed{ \sf{ rate = 38.88 \times {10}^{ - 10} {mol}^{} {L}^{-1} \: {s}^{ - 1} }}$

Answer 2

Answer:

Answer :-

$(1)\to \boxed{ \sf{ rate = 8 \times {10}^{ - 9} \: {mol}^{} {L}^{-1} \: {s}^{ - 1} }} \: \\ \\ (2)\to \boxed{ \sf{ rate = 38.88 \times {10}^{ - 10} {mol}^{} {L}^{-1} \: {s}^{ - 1} }} \:$

To Find :-

(1) Initial rate of reaction for given data .

(2) Calculate rate of reaction after [A] is reduced to 0.06 mol L^{-1}

Explanation :-

Given that

$\to \sf{k = 2.0 \times {10}^{ - 6} {mol}^{ - 2} {L }^{2} {s}^{ - 1} } \\ \\ (1) \sf{ Initial \: rate \: when \: [A] = \: 0.1 \: mol {L}^{ - 1} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{[B] = 0.2 mol {L}^{-1} \: } \\ \\ \to \sf{rate = k[A]{[B]}^{2} \: } \\ \\ \sf{= (2.0 \times {10}^{ - 6} {mol}^{ - 2} {L }^{2} {s}^{ - 1})(0.1mol {L}^{-1}) {(0.2 mol {L}^{-1})}^{2} } \\ \\ \to \sf{rate = 2.0 \times {10}^{ - 6} \times 0.1 \times 0.04 \: \: {mol}^{} {L}^{-1} \: {s}^{ - 1} } \\ \\ \to \boxed{ \sf{ rate = 8 \times {10}^{ - 9} \: {mol}^{} {L}^{-1} \: {s}^{ - 1} }}$

(2) Rate of reaction when [A] reduced to 0.06 mol L^(-1)

So concentration of [A] = 0.1 - 0.06 = 0.04 mol L^(-1)

Therefore concentration of [B] is

[B] = half of [A] = 0.02

→ so available [B] = 0.2 - 0.02 = 0.18

$\to \sf{}rate \: \sf{= (2.0 \times {10}^{ - 6} {mol}^{ - 2} {L }^{2} {s}^{ - 1})(0.06mol {L}^{-1}) {(0.18mol {L}^{-1})}^{2} } \: \\ \\ \to \sf{ rate = 2.0 \times {10}^{ - 6} \times 0.06 \times 0.0324 \: {mol}^{2} {L}^{-1} \: {s}^{ - 1} } \: \\ \\ \to \sf{rate = 0.003888 \times {10}^{ - 6} \: {mol}^{2} {L}^{-1} \: {s}^{ - 1} } \: \\ \\ \to \boxed{ \sf{ rate = 38.88 \times {10}^{ - 10} {mol}^{} {L}^{-1} \: {s}^{ - 1} }}$