For the reaction: 2A + B → A2B The rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−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
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
Do you know any other method of solving this problem ?
For the reaction: 2A + B → A₂B The rate = k[A][B]² with k = 2.0 × 10⁻⁶ mol⁻² L² s⁻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.?
Please reply in below chat box and give use g+ or Like us on Facebook