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For a certain reaction A -> Products with k = 2.0 x 10^-2 s^-1, the initial concentration of A is 1.0 mol L^-1. What concentration of A remains after 100 s if the reaction is first order? (antilog of -0.8684 = 0.135)
A0.500 mol L^-1
B0.270 mol L^-1
C0.135 mol L^-1
D0.050 mol L^-1
Answer & Solution
Correct answer: C. 0.135 mol L^-1
1. First order: $[R] = [R]_0 e^{-kt}$.
2. Exponent $kt = 2.0\times10^{-2}\times100 = 2.0$.
3. $[R] = 1.0\times e^{-2.0}$.
4. $e^{-2.0} = 0.135$, so $[R] = 0.135\ \text{mol L}^{-1}$.
5. Distractor A wrongly treats it as a single half-life; distractor D underestimates by extra decay.
_Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.26_
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