InterviewSolution
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InterviewSolution
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The half life of first order reaction is 990s the initial concentration of the reactant is 0.08mol.dm3 what concentration would remain after 35 minutes |
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Answer» k = 0.693 /t 1/2 = 0.693 /990 s = 7 × 10-4 s-1t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]TT 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 MOL dm-3, t = 35 MIN or 2100 s,t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ?t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ? log10t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ? log10 [A]0/ [A]t = k t /2.303 = 7 × 10-4 s-1 × 2100 s /2.303 = 0.6383t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ? log10 [A]0/ [A]t = k t /2.303 = 7 × 10-4 s-1 × 2100 s /2.303 = 0.6383 [A]0/ [A]t = antilog 0.6383 = 4.35t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ? log10 [A]0/ [A]t = k t /2.303 = 7 × 10-4 s-1 × 2100 s /2.303 = 0.6383 [A]0/ [A]t = antilog 0.6383 = 4.35 Hence, [A]t = [A]0 4.35 = 0.08/ 4.35 = 0.0184 mol dm-3 |
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