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11Alexandr11 [23.1K]
3 years ago
5

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
Chemistry
1 answer:
tangare [24]3 years ago
7 0
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 - <span>initial concentration.
c</span>₁ - <span> concentration at a particular time.
k - </span><span>the rate constant.
t - time.</span>
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