9. For the equibrium reaction shown, K, 10 at reaction at this temperature. Show
ID: 304620 • Letter: 9
Question
9. For the equibrium reaction shown, K, 10 at reaction at this temperature. Show work to support your temperature wi6.225 x10 at 973.15 K. Calculate K for this 10. Consider this reaction: An equilibrium mixture of this reaction at a certain temperature was found to have(Col-0.1055 m, Hal 0.1147 M, and ICH OH]-0.1858 M. What is the value of the equilbrium constant (K) at thuis temperaturer Show work to support your answer. 11. For the synthesis of ammonia: Kc:0.3567 @25. ? A reaction mixture contains 0.6889 M N2, 0.4745 M Ha, and 0.3652 M NHs. Is this reaction at equilibrium? If it is not at equililbrium, indicate the direction in which it much shifts to reach equilibrium. Show work to support your answer. (Hint: Do not use "ICE")Explanation / Answer
Kp, Equilibrium constant based on partial pressure and KC, equilibrium constant based on concentration are related as KP= KC*(RT)deltan, deltan= change in moles of gasesous products
In the reaction NH4NO2 (s) --------->N2(g)+2H2(g)+O2(g)
change in no of moles of gaseous products= moles of gaseous products-moles of gaseous reactants=1+2+1-0= 4
T= 973.15K, R= 0.0821 L.atm/mole.K, Kp =2.25*10-3
KC= KP/(RT)deltan= 5.52*10-11
2. for the reaction , CO(g)+2H2(g)<------>CH3OH(g)
KC= [CH3OH]/[[CO][H2]2= 0.1858/{(0.1055*(0.1147)2}=133.8
3. N2(g)+ 3H2(g)<------->2NH3.
reaction coefficient, Q = [NH3]2/ [N2][H2]3=
=0.3652*0.3652/{(0.6989*(0.4745)3}=1.78
if Q<K, the reaction proceeds towards formation of NH3
if Q=K, the reaction is at equilibrium
if Q>K, the reaction proceeds towards decomposition of NH3
here Q=1.78>KC, so the reaction proceeds backwards, i.e decompostion of NH3
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