Use the References to access important values if needed for this question. Consider the reaction Fe(s) + 2 HCl(aq) FeCl2(s) + H2(g) Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K. ANSWER:
Answer – We are given, reaction –
Fe(s) + 2 HCl(aq) -----> FeCl2(s) + H2(g)
First we need to calculate the ∆Gorxn
We know,
ΔGorxn = ∑ ΔGof of product – ∑ ΔGof of reactant
= [ (ΔGof FeCl2(s)) + (ΔGof H2(g) )] – [ ΔGof Fe (s) + 2*ΔGof HCl(aq)]
= (-302.34 + 0.00) – ( 0.00 + 2*-131.17)
= -42 kJ/mol
Now we know,
ΔGo = -RTlnK
-4.2*104 J = -8.314 J/mol.K * 298.15 K * ln K
So, ln K = -4.2*104 J / -8.314 J.mol-1 .K-1 * 298.15 K
= 16.94
By taking antiln from both side
K = 2.28*107
The equilibrium constant for this reaction at 298.15K is 2.28*107
Use the References to access important values if needed for this question. Consider the reaction Fe(s)...
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