Question

somebody solve that please...​

Answer 1

Explanation:

Given that

[A] = 0.1 mol L−1

[B] = 0.2 mol L−1

k = 2.0 × 10−6 mol−2 L2 s−1

Use the formula of rate of reaction

Rate = k [A][B]2

Plug the values to get initial rate of reaction

Rate = (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1) (0.2 mol L−1)2

= 8.0 × 10−9 L2 mol−2 s−1

Given that value of [A] is change to 0.06 mol−1

Change in concentration of [A]= (0.10 − 0.06) mol L−1 = 0.04 mol L−1

2 mole of A react with 1 mole of B

So that 0.04 mol of A react with =1/2(0.04) = 0.02 mol of B

New concentration of B = 0.2 – 0.02 = 0.18 mol L−1

Use same formula again we get

Rate = k [A][B]2

Plug the values we get

Rate = (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2

= 3.89 mol L−1 s−1

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