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Write Brief Answer · Q25

Q.The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?

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Step 1. k=0.693t1/2=0.69360=0.01155 s−1=1.155×10−2 s−1k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{60}=0.01155\ \text{s}^{-1}=1.155\times10^{-2}\ \text{s}^{-1}.

Step 2. 180 s÷60 s (per half life)=3180\ \text{s}\div60\ \text{s (per half life)}=3 half lives have elapsed.

Step 3. After n half lives, the fraction of A remaining is (1/2)n(1/2)^n; here n=3n=3: (1/2)3=1/8=0.125=12.5%(1/2)^3=1/8=0.125=12.5\%. …

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