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Choose the Best Answer · Q1

Q.For a first order reaction A→BA \rightarrow B, the rate constant is x min−1x\ \text{min}^{-1}. If the initial concentration of A is 0.01 M, the concentration of A after one hour is given by the expression.

(a) 0.01 e−x0.01\ e^{-x}
(b) 1×10−2 e−60x1\times10^{-2}\ e^{-60x}
(c) 1×10−2(1−e−60x)1\times10^{-2}(1-e^{-60x})
(d) none of these
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✓ Free question

Step 1. The exponential form of the first order integrated rate law is [A]=[A]0e−kt[A]=[A]_0e^{-kt} (directly equivalent to ln⁡([A]0/[A])=kt\ln([A]_0/[A])=kt).

Step 2. Here [A]0=0.01 M=1×10−2[A]_0=0.01\ \text{M}=1\times10^{-2} M, k=x min−1k=x\ \text{min}^{-1}, and t=1t=1 hour =60=60 min.

Step 3. Substituting: [A]=(1×10−2)e−x×60=1×10−2 e−60x[A]=(1\times10^{-2})e^{-x\times60}=1\times10^{-2}\ e^{-60x}.

✓Final answer

(b) 1×10−2 e−60x1\times10^{-2}\ e^{-60x}

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