lecture6

# 5 kj h 2858 kj h 31196 kj 2xcgraphite o2g co2g

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Unformatted text preview: iamond) + O2(g) ∆H = +395.4 kJ C(graphite) Æ C(diamond) ∆H = +1.9 kJ Calculate ∆H for the reaction 2C(graphite) + 3H2(g) Æ C2H6(g) from the following data C(graphite) + O2(g) Æ CO2(g) H2(g) + 1/2O2(g) Æ H2O(l) 2C2H6(g) + 7O2(g) Æ 4CO2(g) + 6H2O(l) ∆H = -393.5 kJ ∆H = -285.8 kJ ∆H = -3119.6 kJ 2x(C(graphite) + O2(g) Æ CO2(g)) 2C(graphite) + 2O2(g) Æ 2CO2(g) 3x(H2(g) + 1/2O2(g) Æ H2O(l)) 3H2(g) + 3/2O2(g) Æ 3H2O (l) -1/2 x( 2C2H6(g) + 7O2(g) Æ 4CO2(g) + 6H2O(l)) 2CO2(g) + 3H2O (l) Æ C2H6(g) + 7/2O2(g) ∆H = 2 x (-393.5 kJ) ∆H1 = -787.0 kJ ∆H = 3x(-285.8 kJ) ∆H2 = -857.4 kJ ∆H = -1/2 x (-3119.6 kJ) ∆H3 = +1,559.8 kJ 2C(graphite) + 3H2(g) Æ C2H6(g) ∆H = ∆H1 + ∆H2 + ∆H3 = -84.6 kJ 6-5 Chapter 6 Thermochemistry 6.4 Standard Enthalpies of Formation Standard Enthalpy of Formation, ∆H°f is the enthalpy change for the formation of 1 mole of a substance from the elements in their standard states. Standard state: gases at 1 atm pressure, solution at 1 M, pure liquid or solid elements: stable form at 1 atm and 25° C by definition, ∆H°f for an element in its standard state is 0 any reaction can be considered to take the path then reactants Æ elements Æ products ∆Hrxn = ∑∆H°f(products) - ∑∆H°f(reactants) 2C2H2(g) + 5O2(g) Æ 4CO2(g) + 2H2O(g) 2(g) + 5(g) Æ 4(g) + 2(g) ∆H = 4(∆Hf(CO2)) + 2(∆Hf(H2O)) -2(∆Hf(C2H2)) -5(∆Hf(O2)) ∆H = 4(-393.5 kJ) + 2(-241.8 kJ)) -2(+226.7 kJ) -5(0 kJ) ∆H = -1574.0 kJ - 483.6 kJ - 453.4 kJ - 0 kJ = -2511.0 kJ KClO3 + 3PCl3 Æ 3POCl3 + KCl ∆H = 3(∆Hf(POCl3)) + (∆Hf(KCl)) -(∆Hf(KClO3)) -3(∆Hf(PCl3)) ∆H = 3(-597.0 kJ) + (-435.9 kJ)) -(-391.4 kJ) -3(-306.4 kJ) ∆H = -916.3 kJ 6.5 Present Sources of Energy 6.6 New Energy Sources 6-6...
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## This note was uploaded on 04/05/2014 for the course CHEM 1211 taught by Professor Jackduff during the Spring '13 term at SPSU.

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