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The aqueous reaction A -> B + C is first order. The progress of the reaction is followed by titrating samples with a reagent R at time t and at t = infinity, consuming n1 and n2 moles of R respectively. If A, B, and C each react with R, and their n-factors with R are in the ratio 1:2:3 respectively, express the rate constant k in terms of t, n1, and n2.
- k = (1/t) * ln(n2 / (n2 - n1))
- k = (1/t) * ln(2*n2 / (n2 - n1))
- k = (1/t) * ln(4*n2 / (n2 - n1))
- k = (1/t) * ln(4*n2 / (5*(n2 - n1)))
Correct answer: k = (1/t) * ln(4*n2 / (5*(n2 - n1)))
Solution
Let initial moles of A = a, x = moles reacted at time t. At t=inf all A->B+C: n2 = 2a + 3a = 5a, so a = n2/5. At time t: n1 = 1*(a-x) + 2x + 3x = a + 4x, so x = (n1 - a)/4 = (n1 - n2/5)/4. Remaining A = a - x = n2/5 - (n1-n2/5)/4 = (4n2/5 - n1 + n2/5)/4 = (n2 - n1)/4. First-order: k = (1/t)*ln(a/(a-x)) = (1/t)*ln((n2/5)/((n2-n1)/4)) = (1/t)*ln(4n2/(5(n2-n1))).
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