( d ) 2CH4(g)- C,H6(g) + H2(g) Equilibrium Constants and G 18.97 Calculate the v
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Question
( d ) 2CH4(g)- C,H6(g) + H2(g) Equilibrium Constants and G 18.97 Calculate the value of the thermodynamic equilibrium constant for the following reactions at 25 °C. (Refer to the data in Appendix C.) (a) 2PC13(g) + O2(g)- 2POCI,(g) (b) 2SO (g2SO2(g 18.98 Calculate the value of the the modyuamic equilibrium constant for the following reactions at 25 °C. (Refer to the data in Appendix C) Gohjo,--105.6 kj mol-1 (a) NHg +20,)2NOg+2H,0( (b) N,H,(g) + 6H202(g)-2N02(g) + 8H20(g) 3.99 The reaction NO2(g) +NO(gN,O(g +0(3 67 kI A 1 00 L, reaction vessel atExplanation / Answer
Answer – 18.98 a) We ae given the reaction
N2H4(g) + 2O2(g) -----> 2NO(g) + 2H2O(g)
First we need to calculate Go rxn is ,
Go rxn = sum of the Go product – sum of the Go reactant
= [2*Go NO(g) + 2*Go H2O(g) ] – [Go N2H4 (g)+ 2* Go O2(g)]
= ( 2*86.57 + 2*-228.59 ) - (159.2 + 2*0.00)
= -443.24 kJ/mol
We know the formula,
Go = -RT ln K
So, Go in J
1 kJ = 1000 kJ
So, -443.24 kJ = ?
= -4.43*105 J/mol
Now plugging the values in the formula,
-4.43*105 J/mol = - 8.314J.mol-1.K-1 * 298 K * ln K
ln K = -4.43*105 J/mol / - 8.314J.mol-1.K-1 * 298 K
ln K = -178.9
now taking antiln from both side
K = 2.015*10-78
ab) We ae given the reaction
N2H4(g) + 6H2O2 (g) -----> 2NO2(g) + 8H2O(g)
First we need to calculate Go rxn is ,
Go rxn = sum of the Go product – sum of the Go reactant
= [2*Go NO2(g) + 8*Go H2O(g) ] – [Go N2H4 (g)+ 6*Go H2O2(g)]
= ( 2*51.30 + 8*-228.59 ) - (159.2 + 6*-105.6)
= -1251.7 kJ/mol
We know the formula,
Go = -RT ln K
So, Go in J
1 kJ = 1000 kJ
So, -1251.7 kJ = ?
= -1.251*106 J/mol
Now plugging the values in the formula,
-1.251*106 J/mol = - 8.314J.mol-1.K-1 * 298 K * ln K
ln K = -1.251*106 J/mol / - 8.314J.mol-1.K-1 * 298 K
ln K = -505.2
now taking antiln from both side
K = 3.85*10-220
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