Standard Thermodynamic Quantities for Selected Substances at 25 C Part A For the

Question

Standard Thermodynamic Quantities for Selected Substances at 25 C

Part A

For the reaction calculate Hrxn at 25 C.
2CH4(g)C2H6(g)+H2(g)

Express your answer using one decimal place.

64.6

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Correct

Part B

For the reaction calculate Hrxn at 25 C.
2NH3(g)N2H4(g)+H2(g)

Express your answer using one decimal place.

187.2

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Correct

Part C

For the reaction calculate Hrxn at 25 C.
N2(g)+O2(g)2NO(g)

Express your answer using one decimal place.

182.6

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Correct

Part D

For the reaction calculate Hrxn at 25 C.
2KClO3(s)2KCl(s)+3O2(g)

Express your answer using one decimal place.

-77.6

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Correct

Part K

For the reaction in part C calculate Grxn at 25 CC.

Express your answer using one decimal place.

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Incorrect; Try Again; 7 attempts remaining

Part L

For the reaction in part D calculate Grxn at 25 C.

Express your answer using one decimal place.

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Substance Hf(kJ/mol) S(J/molK) CH4(g) 74.6 186.3 C2H6(g) 84.68 229.2 H2(g) 0 130.7 N2(g) 0 191.6 NH3(g) 45.9 192.8 N2H4(g) 95.4 238.5 NO(g) 91.3 210.8 O2(g) 0 205.2 KCl(s) 436.5 82.6 KClO3(s) 397.7 143.1

Explanation / Answer

Ans

Part K = 175.21 kJ

Part L = -225.06 kJ

Exp :

Part K : N2 (g) + O2 (g) -------> 2NO (g) , dH = 182.6 kJ/mol , T = 25 + 273.15 K

dS = So ( products) - So (reactants)

dS = [ 2 x 210.8] -[ 191.6 + 205.2] = 24.8 J/mol K

Therefore dG = dH -TdS

dG = 182600 J/mol - [298.15 K x 24.8 J/mol K]

dG = 175.21 kJ/mol

Part L ; 2KCLO3 (s) ------> 2KCl(s) + 3O2

dH = -77.6 kJ/mol , T = 25 + 273.15 K

dS = So ( products) - So (reactants)

dS = [ (2 x 82.6) + (3 x 205.2)] -[2x143.1 ] = 494.6 J/mol K

Therefore dG = dH -TdS

dG = -77600 J/mol - [298.15 K x 494.6 J/mol K]

dG = -225.06 kJ/mol

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