± Pressure-Based versus Concentration-Based Equilibrium Constants Part A For the
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Question
± Pressure-Based versus Concentration-Based Equilibrium Constants
Part A
For the reaction
3A(g)+3B(g)C(g)
Kc = 32.8 at a temperature of 377 C .
Calculate the value of Kp.
Express your answer numerically.
Part B
For the reaction
X(g)+3Y(g)2Z(g)
Kp = 1.95×102 at a temperature of 369 C .
Calculate the value of Kc.
Express your answer numerically.
The Reaction Quotient
The following reaction was carried out in a 3.25 L reaction vessel at 1100 K:
C(s)+H2O(g)CO(g)+H2(g)
If during the course of the reaction, the vessel is found to contain 9.25 mol of C, 14.6 mol of H2O, 3.40 mol of CO, and 8.20 mol of H2, what is the reaction quotient Q?
Enter the reaction quotient numerically.
Explanation / Answer
Part A :
3A(g)+3B(g)C(g) : Kc = 32.8
Kp = Kc x (RT)dn
Where dn = change in number of moles
= Number of moles of gaseous products - number of moles of gaseous reactants
= 1 - ( 3+3)
= -5
R = gas constant = 8.314 J/mol-K
T = Temperature = 377 oC = 377+273 = 650K
Plug the values we get Kp = Kc x (RT)dn
= 32.8 x ( 8.314x650)-5
= 7.12x10-18
Part B :
X(g)+3Y(g)2Z(g) :Kp = 1.95×102
Kp = Kc x (RT)dn
Where dn = change in number of moles
= Number of moles of gaseous products - number of moles of gaseous reactants
= 2 - (1+3)
= -2
R = gas constant = 8.314 J/mol-K
T = Temperature = 369 oC = 369+273 = 642 K
Plug the values we get Kc = Kp x (RT)-dn
= 1.95×102 x ( 8.314x642)-(-2)
= 5.55x105
Part C :
Number of moles of C = moles / volume = 9.25 mol / 3.25 L = 2.85 M
Number of moles of H2O = moles / volume = 14.6 mol / 3.25 L = 4.49 M
Number of moles of CO = moles / volume = 3.40 mol / 3.25 L = 1.05 M
Number of moles of H2 = moles / volume = 8.20 mol / 3.25 L = 2.52 M
C(s)+H2O(g)CO(g)+H2(g)
Reaction quotient , Q = ([CO][H2])/([C][H2O])
= (1.05x2.52)/(2.85x4.49)
= 0.21
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