A common reaction used to study reaction kinetics is the dissolution of sodium thiosulfate in hydrochloric acid. This process is monitored by cloudiness in the container for the following reaction: 2HCl(aq) + Na2S2O3(aq) → S(s) + SO2(g) +2NaCl(aq) + H2O(l) b) Suppose 1 M Na2S2O3 increased the rate of the reaction but 1 M HCl did not affect the reaction rate. Write 3 - 4 sentences to describe what these results tell about the rate law and the mechanism for the reaction. Explain your reasoning. I have no idea how to do this question, so if someone could give an explanation as to why Na2S2O3 increased the reaction rate but HCL had no affect, it would be very helpful.

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A common reaction used to study reaction kinetics is the dissolution of sodium thiosulfate in hydrochloric acid. This process is monitored by cloudiness in the container for the following reaction: 2HCl(aq) + Na2S2O3(aq) → S(s) + SO2(g) +2NaCl(aq) + H2O(l) b) Suppose 1 M Na2S2O3 increased the rate of the reaction but 1 M HCl did not affect the reaction rate. Write 3 - 4 sentences to describe what these results tell about the rate law and the mechanism for the reaction. Explain your reasoning. I have no idea how to do this question, so if someone could give an explanation as to why Na2S2O3 increased the reaction rate but HCL had no affect, it would be very helpful.

Chemistry
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To answer this question doesn't actually matter \(why\) one reactant increases the rate and one doesn't, it's what the observations \(mean\).
the rate law is the mathematical equation that describes the factors (like concentrations of reactants) that make a reaction run faster or slower
@JFraser Alright. So does this mean that the dissolution of sodium thiosulfate in hydrochloric acid is a multi-step mechanism, where sodium thiosulfate is part of the rate-determining step, or the slow step, and HCL is part of the fast step and thus, increasing its concentration has little to no affect on the reaction rate?

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that's what I would infer about this problem, yes
Thank you!
YW

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