What is the rate, mol s^-1, at the instant when [OH] = 1.7 × 10^-8 M and [H2S] = 1.0 × 10^-8 M and t — Chemical Kinetics Chemistry Question
Question
What is the rate, mol s^-1, at the instant when [OH] = 1.7 × 10^-8 M and [H2S] = 1.0 × 10^-8 M and the volume of the reacting system is 0.1 litre?
Answer: B
💡 Solution & Explanation
Rate (in mol s^-1) = Rate (in mol L^-1 s^-1) × Volume (in L). Rate = 8.71 × 10^-7 × 0.1 = 8.71 × 10^-8 mol s^-1. Therefore, correct answer is B.
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