Question 2

For the reaction:

2A + B → A₂B

the rate = k[A][B]²with k= 2.0 x 10^-6mol^-2L²s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.

Answer

The initial rate of the reaction is

Rate = k [A][B]²

= (2.0 × 10^{- 6}mol^{- 2}L²s^{- 1}) (0.1 mol L^{- 1}) (0.2 mol L^{- 1})²

= 8.0 × 10^{- 9}mol^{- 2}L²s^{- 1}

When [A] is reduced from 0.1 mol L^{- 1}to 0.06 mol^{- 1}, the concentration of A reacted = (0.1 - 0.06) mol L^{- 1} = 0.04 mol L^{- 1}

Therefore, concentration of B reacted= 1/2 x 0.04 mol L^-1 = 0.02 mol L^{- 1}

Then, concentration of B available, [B] = (0.2 - 0.02) mol L^{- 1}

= 0.18 mol L^{- 1}

After [A] is reduced to 0.06 mol L^{- 1}, the rate of the reaction is given by,

Rate = k [A][B]²

= (2.0 × 10^{- 6}mol^{- 2}L²s^{- 1}) (0.06 mol L^{- 1}) (0.18 mol L^{- 1})²

= 3.89 mol L^{- 1}s^{- 1}

4 Comment(s) on this Question

dinchak pooja 2019-09-21 05:22:19

oohhhooo AHhhaa

Shubhangee 2019-09-10 17:31:11

How come we halved the concentration of A to find the new one of B

Yyy 2019-06-05 15:47:16

Ohho

Aynal Hoque 2018-10-23 11:05:31

Second part of this answer\'s method is correct but answer is wrong...it will be = 3.888*10^-9 mol-2L2s-1