Question

QUESTION 17 Suppose the surface-catalyzed hydrogenation reaction of an unsaturated hydrocarbon has a rate constant of 0681 M min. The reaction is observed to follow zero order kinetics. If the initial concentration of the hydrocarbon is 8.30 M, what is the half-life of the reaction in seconds? "Please report 3 significant figures Numbers only, no unit. No scientific notation QUESTION 18 Which of the following expressions is the correct equilibrium constant expression for the reaction below 3 F60(0) + CO2(g) Fe3O4(8) CO() [Coy (CO2] 100][00] (Fe, Ojco [FO][CO] [FeoPCO) [Fe20][CO] 1/CO

Answer 1

17.) Zero order reaction means rate of reaction does not change due to change in concentration of the reactant.

DA/Dt = -k

Half life of a reaction is the time taken by reaction to consume half of the initial concentration of reactant.

So, at t = T1/2 , At = Ao / 2 , where Ao is initial concentration of the reactant.

On integrating above equation, we get

At - Ao = -kt

For t = T1/2

  T1/2 = Ao / 2k

Given, k = 0.681 M / min

Ao = 8.30 M

So, T1/2 = 8.30 /2*0.681

T1/2 = 366 s

18.) Pure solid and liquid are not included in equilibrium constant expression. This is because they affect the reactant amount at equilibrium in the reaction, so they are disregarded and kept at 1.

Given reaction:

  

3Fe(s) + CO2(g) = Fe3O4(s) + CO(g)

K = [CO] / [CO2]

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