A possible mechanism for a reaction that takes place in solution is Step (1): Ce

Question

A possible mechanism for a reaction that takes place in solution is Step (1): Ce^4+ + Mn^2+ rightarrow Ce^3+ + Mn^3+ slow Step (2) Ce4^+ + Mn^3+ rightarrow Ce^3+ + Mn^4+ fast Step (3) Mn^4+ + Tl^1+ rightarrow Mn^2+ + Tl^3+ fast what is the overall reaction what is the predicted rate law which species is (are) intermediates, and which species is the catalyst? A mixture with [N_2O_4] = 0.90 and [NO_2] = 0.134 at 25 degree C is at equilibrium as follows: N_2O_4(g) rlhar 2NO_2(g) What is K_c for the reaction, and how can you drive the reaction to the right without changing the temperature? The equilibrium constant, K_c, for the reaction: C(s) + CO_2(g) rlhar 2 CO(g) is 0.113 at 1100 K. What is the concentration of CO in a mixture at equilibrium if the CO_2 concentration is 0.020 M, and how would you 'drive' the reaction to completion at 1100 K? Answer (a) and (b) If K_c = 7.5 times 10^-9 at 1000 K for the reaction: N_2(g) + O_2(g) rlhar 2 NO(g) what is K_p at 1000 K for the reaction, and; what is K_c for 2 NO(g) rlhar N_2(g) + O_2(g) at 1000 K?

Explanation / Answer

Question 5.

a) add the reaction 1 + 2+ 3 to get the overhall reaction, adding 1 and 2 and 3 we get

2Ce+4 + Tl+1 ------------------------> 2Ce+3 + Tl+3

b) The slowest step is the rate determining step , Hence the step 1 is the rate determining step. So rate depend on both the reactants as we have two reactants in rate determining step

SO rate = k [Ce+4][Mn+2] , Hence the reaction is of order 2.

c) Mn+3  and Mn+4 act as an intermediates in the reaction and are produced and consumed within the reaction steps.

d) Here Mn+2 act as a catalyst because it make the reaction faster by further changes and reactions and is generated at the end of the reaction.

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