The reaction of thioacetamide with water is shown by the equation below: CH3C(S)NH2(aq) + H2O → H2S(aq) + CH3C(O)NH2(aq). The rate of reaction is given by the rate law: Rate = k[H3O^+][CH3C(S)NH2]. Co

Question

The reaction of thioacetamide with water is shown by the equation below: CH3C(S)NH2(aq) + H2O → H2S(aq) + CH3C(O)NH2(aq). The rate of reaction is given by the rate law: Rate = k[H3O^+][CH3C(S)NH2]. Consider 1 l of solution that is 0.20 M in CH3C(S)NH2 and 0.15 M in HCl at 25°C. For each of the changes listed below, the number of changes that decreases the rate of reaction is
(i) A 4.0 g sample of NaOH is added to the solution.
(ii) 500 ml of water is added to the solution.
(iii) The 0.15 M HCl solution is replaced by 0.15 M acetic acid solution.
(iv) The reaction is carried out at 35°C instead of 25°C.

Prashant Kotian · Accepted Answer

Correct answer: 3. (i) Adding NaOH neutralizes some H3O^+, lowering its concentration and the rate (decreases). (ii) Dilution by adding water lowers concentrations of both reactants (decreases). (iii) Acetic acid is weak, providing much less H3O^+ than HCl (decreases). (iv) Increasing temperature generally increases the rate constant and the rate (increases). Thus, 3 changes decrease the rate. Therefore, correct answer is 3.

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