For the following reaction, calculate H° rxn (kJ/mol) using H° f CH 4 (g) + 4 Cu

Question

For the following reaction, calculate H°rxn (kJ/mol) using H°f

CH4 (g) + 4 CuO (s) CO2 (g) + 2H2O (l) + 4Cu (s)

Given H°f :

CH4 (g)   –74.9 kJ/mol

CO2 (g)   –393.5 kJ/mol

CuO (s)   –145.6 kJ/mol

H2O (l)    –285.8 kJ/mol

____ Units ____

Can you please explain what you did? I'm having trouble understanding this problem.

What is AH f (kJ/mol) for N2H4 (I) given the following information (3 sig figs): 2 4 (l) 2 NO 2 (g) 2 2 (l) AH 508.7 kJ 2(g) AHof (kJ/mol): NoOual +81.1 H2Oa, -241.8 H2O -285.8 HF -271.1 NOway +90.3 NOoua +34.3 N204(a) +9.2 PC13(IV -319.7 CCI40 -135.4; CH40g -74.8; CH3OHO) -238.7; C2H50Ho --277.7; coCo --110.5 CO2(g) -393.5; C2H20g) +226.7; C 2H6(g) --84.7 2H4(g) +52.3 3H8(g) -103.8 4H10(g) -888.0 Units

Explanation / Answer

dH rxn = dH ( products) - dH ( reactants)

dH rxn = dH ( CO2) + 2dH (H2O) + 4 dH ( Cu) - dH ( CH4) - 4 dH ( CuO)

dH rxn = -393.5 KJ/mol + 2( -285.8 KJ) + 4 ( 0) - (-74.9 KJ) - 4 ( -145.6 KJ)

           = - 307.8   KJ/mol

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