I2(aq) + 2S2O32-(aq) rightarrow 2I-(aq) + S4O62-(aq) How much did the temperatur
ID: 891984 • Letter: I
Question
I2(aq) + 2S2O32-(aq) rightarrow 2I-(aq) + S4O62-(aq) How much did the temperature change during the course of the experiment? Increasing temperature speeds up a reaction. Decreasing temperature slows down a reaction. If there were a temperature change of over 2 degrees. try Interpret if variations in your data are consistent with the temperature change. The rate law for the generic reaction rate = k[A)x(B]y could be verbally described as: The reaction is first order in substance A. first order in substance B. and second order overall. Use this style to describe the rate law for the reaction you used in tins experiment. If the order of a reaction is the same as its stoichiometry. the order of each reactant is the same number as the coefficient of that reactant in the balanced equation. When this is the case, it is possible (but not necessary) that the reaction takes place in one step. If the order of a reaction is different than its stoichiometry. then the reaction must take place in steps. For the reaction used in this experiment. show how the rate law (equation 1 on page 33) would look if the reaction took place in one step:Explanation / Answer
1) As the temperature of the experiment is increased the rate of reaction also goes up. Going with the general principle, the rate of reaction doubles with every 10 oC rise in temperature. Temperature increases rate of coillison between the reactants which eventually leads to faaster reaction.
2) The rate law for the given reaction would be,
rate = k[I2][S2O3^2-]
where,
k = rate constant
overall order of reaction = 1 + 1 = 2
3) If the order is follows the stoichiometry of the reaction, the rate law for this reaction can be written as,
rate = k[I2][S2O3^2-]^2
Overall order for this reaction = 1 + 2 = 3
this is when the reaction takes place in one step.