Question

1) At 298 K, an equilibrium mixture contains P H2=0.958 atm, P I2= 0.877 atm and P HI=0.020 atm. Please try and answer all!

You must show all your work to receive credit. Watch sig figs! Consider the reaction: 1) At 298 K, an equilibrium mixture contains P H2-0.958 atm, P ½=0.877 atm, and P HI-0.020 atm. Calculate the value of the equilibrium constant Kp? a) (o, o2) 0.938x o.87? b) And what is the value of the equilibrium constant Kc? Predict whether the reaction is spontaneous or non-spontaneous at 298 K. Explain yo reasoning. c) d) Determine the value of ΔG° of the reaction under the same conditions. e) What does the calculated value of Does this agree with what you proposed in part c? AG tell you about the spontaneity of the reacti

Answer 1

Stuen Mulg)+14) = 2414) 1. Pry= 0.958 atm Pq = 0.877 als PHI=0.020 atm (a) Equilibrium constant sky Kp = (PHL) - SDS- Kp = (0.0202 [0-958) (0.877) Kp = 4076 x104] (6) Equilibrium constant ake As we know on Kp = Ke (RT)" here ana change in number of moles sn=2=(1+1)=0 Kp = Kc (RT)° [Kp = kg = 40763109 (C) The reaction is non-spontaneous at 298k, because - Ep and Ke are both less than 1. for spontaneous Reaction top and kee must be greater than I. (d) OG = AG® +RT enke (for equilibrium AG=0) OG = -RT enke AG°= -0.008314 KJ moy?x7x298 K lm (4.76x109 CS Camscan y G=-18953 kJ molt

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