Solution:
Given, [A]=0.1 mol L-1, [B]=0.2 mol L-1, k = 2.0×10-6 mol-2 L2 s-1Rate =k[A][B]2
= (2.0x 10-6 ) x (0.1 mol L-1) x (0.2 mol L-1) 2
= 8.0 x 10-9 mol-2 L2 s-1.
The concentration of A is reduced to (0.1 – 0.06) = 0.04 mol L-1
.Now according to the equation 1 mole of A reacts with 2 mole of B.
= (2.0x 10-6 ) x (0.1 mol L-1) x (0.2 mol L-1) 2
= 8.0 x 10-9 mol-2 L2 s-1.
The concentration of A is reduced to (0.1 – 0.06) = 0.04 mol L-1
So, 0.04 mol of A react with B, hence the no. of moles of B reacted will be 2 x 0.04 = 0.08.
Now, concentration of B will be (0.2 – 0.08) = 0.12 mol L-1
i.e., [A]=0.06 mol L-1 and [B] = 0.12 mol L-1
i.e., [A]=0.06 mol L-1 and [B] = 0.12 mol L-1
Put the new concentrations in the rate expression given, we get
R = 2.0 x 10-6 x 0.06 x (0.12)2
= 1.72 x 10-9 mol-2 L2 s-1
Pingback: From 200 mg of CO2, 10^21 molecules are removed. How many moles of CO2 are left? - CG's Chemistry Solutions