## Answers

Ane) Given Data is a) Kc 4.73x105. - Kf = 8.38x1035 kg 2 1.97 xjö? 57 Forward rate constant = 3.52x10 st.

We already know the formula Given problem is form where ke= equilibrium constant forward reaction take const Kf = kra reverse reaction rake constant . . a Adding rate Catalyst increases the forward constant kg = 3.52x10 st Sub stilve the above equ to find out kg 4.73x105 a 3.52x105 st - kr CSScanned with CamScanner

We have to find ka value After adding a Со Catalyst. : : . Kor= 3.52x16 51 4.73 xics Kor= 3.52 Kor: 0.74418 st b) Given.

Data is, Sealed vessel at 778°c Hi cg) + I, 9) = QHI(9) Concentration of [14] - 3.15m Concentration of [14] - 2.85min Concentration = 3.75M Equilibrium concentration of Il = 0.0700 M Hacm) [I.].m. | [HI],m Initial I 2.85m 3.75 M change +2x Swilibriomand (3.75-2) 2.85 Dgium 1.75-2 SScanned with Camscanner (.85-x)/ +2x

Since. [12] combination = 0.0700 = 0.0400 = 2.85-X 0.07 - 2.85-7 x 2.85-0.07 The following concentrations are get 3.75-X = 3.75 - 2.78 = 0.97 Mon(moly 0.97 M q? 0.0700 ml - 22. - 2x 2.78 M 2 5.56M | [HI], 5.5cm 1 5.56M CSscanned with CamScanner

. Let or find out Equilibrium constant 0 kc from Ice table (5.56) Kc = (0.97)(0.07) 30.9136 kez 0.0679 1 Kc = 455.2812 ..

.
Equilibrium constant Ikea 455.28 Kc = 455.28 SScanned with CamScanner

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