The reaction 2X →products is a second-order reaction, and the rate constant is 8.8 * 10-3 1/M·s. If the initial concentration of X is 3.00 M, how many seconds does it take for the concentration of X to drop to 0.700 M?

A) 124 s

B) 261 s

C) 0.0096 s

D) 200 s

Question

Myla

Chemistry

10 November, 10:44

The reaction 2X →products is a second-order reaction, and the rate constant is 8.8 * 10-3 1/M·s. If the initial concentration of X is 3.00 M, how many seconds does it take for the concentration of X to drop to 0.700 M?

A) 124 s

B) 261 s

C) 0.0096 s

D) 200 s

+3

Answers (1)

Know the Answer?

Jayce Norman · Answer 1

Answer is: A) 124 s.

c₀ = 3 mol/L.

c₁ = 0,700 mol/L.

k = 8,8·10⁻³ 1/M·s.

Integrated second order rate law is: 1/c₁ = 1/c₀ + k·t.

k·t = 1/0,700 - 1/3.

0,0088·t = 1,095.

t = 1,095 : 0,0088.

t = 124 s.

c₀ - initial concentration.

c ₁ - concentration at a particular time.

k - the rate constant.

t - time.

The reaction 2X →products is a second-order reaction, and the rate constant is 8.8 * 10-3 1/M·s. If the initial concentration of X is 3.00 M, how many seconds does it take for the concentration of X to drop to 0.700 M?A) 124 sB) 261 sC) 0.0096 sD) 200 s

The reaction 2X →products is a second-order reaction, and the rate constant is 8.8 * 10-3 1/M·s. If the initial concentration of X is 3.00 M, how many seconds does it take for the concentration of X to drop to 0.700 M?

A) 124 s

B) 261 s

C) 0.0096 s

D) 200 s