Question

1. The rate constant for a certain reaction is k = 7.10×10⁻³ s⁻¹ . If the initial reactant concentration was 0.350 mol L⁻¹, what will the concentration be after 19.0 minutes?

2. A zero-order reaction has a constant rate of 3.70×10⁻⁴ mol L⁻¹ s^−1. If after 40.0 seconds the concentration has dropped to 9.00×10⁻² mol L⁻¹, what was the initial concentration?

Answer 1

1)

Time = 19.0 x 60=1140 s

Using formula as follows:

ln (x/ 0.350) = - 7.10x 10^-3 x 1140= - 8.094

e^-8.094=0.000305366 = x / 0.350
x = 0.0001068 M
Therefore, answer is 0.0001068 M.

2)

A =A0 - kt

A0 = 9.00×10−2 + (3.70×10−4)*40

A = 0.1048 mol/L

Therefore, answer is 0.1048 mol/L