Using data from Appendix 4, calculate delta H degree, delta S degree , and K (at

Question

Using data from Appendix 4, calculate delta H degree, delta S degree , and K (at 298 K) for the synthesis of ammonia by the Haber process. (Use the values for AG from Appendix 4 when calculating K.) N2(g) + 3 H2(g) rightarrow 2 NH3(g) Calculate delta G for this reaction under the following conditions (assume an uncertainty of plus minus 1 in all quantities). Assume that delta H degree and delta S degree do not depend on temperature. T = 273 K, PN2 = PHz = 200 atm, PNh3 = 50. atm T = 283 K, PNz = 160 atm, PHz = 540 atm, PNh3 = 200. atm T = 197 K, Pm2 = 40. atm, Ph2 = 248 atm, PNh3 = 33 atm T = 622 K, PNz = 20. atm, Ph2 = 119 atm, PNh3 = 14 atm

Explanation / Answer

dH = 2dH (NH3) = 2x(-46) = -92 KJ, dS = 2dS(NH3)-3dS(H2)-dS(N2) , = 2(192)-3(131)-193 = -200 Joules/K, dG = dH-TdS = -92000 -298(-200) = -32400 J, dG = -RT ln K , -32400 = -8.314 x 298 lnK , K = 4.778 x10^ 5 , a) dG = -8.314 x 273 lnK , K = [NH3]^2/[N2][H2]^3 = 50^2/(200x200^3), K = 1.56 x10^ -6 , dG = -(8.314 x298 x2.3/1000)xlog(1.56 x10^ -6) = 33.089 KJ , b) K = (200^2)/(540^3)(160) = 1.587 x 10^ -6 , dG = (-8.314 x283 x2.3/1000) log(1.587 x10^ -6) =31.384 KJ c) K = 33^/(248^3 x 40) = 1.78 x 10^ -6 , dG = (-8.314 x197 x2.3/1000) log(1.78 x10^ -6) = 21.659 KJ d) K = 14^2/(119^3 x 20) = 5.8 x10^ -6 , dG = (-8.314 x622 x2.3/1000) log(5.8 x10^ -6) = 62.284 KJ

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