show all work please NH4F(s) NH3(g) + HF(g) Determine (in kJ) for this reaction
ID: 780562 • Letter: S
Question
show all work please
NH4F(s) NH3(g) + HF(g) Determine (in kJ) for this reaction at 170.76 K. Assume and S degree do not vary as a function of temperature. Report your answer to two decimal places Determine the equilibrium constant for this reaction at 170.76 K. Report your answer to three significant figures in scientific notation. If the partial pressure of HF is 3.04 atm and the partial pressure of NH3 is 1.1 atm, determine Delta G (in kJ) for this reaction at 170.76 K. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ)Explanation / Answer
1) dH rxn = -272.55-45.94-(-464) = 145.51 KJ,
dS = 173.78+192.77-72 = 294.55 J = 0.29455 KJ,
dG0 = 145.51-(170.76)(0.29455) = 95.2 KJ
2) dG0 = -RT ln K , 95200 = -8.314 x170.76 ln K ,
K = 7.48 x10^ -30
3) Kp = pNH3 xpHF = 3.04 x1.1 = 3.344,
Kc = Kp/(RT)^dn dn = moles of gas product-moles of gas reactanst = 2,
Kc = 3.344/(0.0821x170.76)^2 = 0.017
dG = dGo + RT ln K = 95.2 + (8.314x170.76/1000) ln(0.017)
= 89.42 KJ
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