Jump to... Arth 29/2017 11:55 PM 5.6/13 10/29/2017 02:06 PM Gradeb Print Calcuia

Question

Jump to... Arth 29/2017 11:55 PM 5.6/13 10/29/2017 02:06 PM Gradeb Print CalcuiatorPeriodic Table on 1 of 13 Map Sapling Learning At a given temperature, the elementary reaction AB in the forward direction is first order in A with a rate constant of 3.20 102 s-1. The reverse reaction is first order in B and the rate constant is 630 x 102s-1 What is the value of the equilbrium constant for the reaction AB at this temperature? Number What is the value of the equiliborium constant for the reaction BA at this temperature? 20 oO F F7 F8 F5 F6

Explanation / Answer

1)

Kc = Kf/Kb

= (3.20*10^-2)/(6.30*10^-2)

= 0.508

Answer: 0.508

2)

This reaction is reverse of reaction given in part 1

So,

Kc2 = 1/Kc1

=1/0.508

= 1.97

Answer: 1.97

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