1.

With the help of thermochemical equation, calculate ΔfH° at 298 K for the following reactions: C(graphite) + O2(g) → CO2(g) ; ΔfH° = -393.5kJ/mol H2(g) + 1/2O2(g) → H2O(l) ; ΔfH° = -285.8kJ/mol CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ; ΔfH° = +890.3kJ/mol

Answer»

C(graphite) + O2(g) → CO2(g) ; ΔfH° = -393.5 kJ/mol …(1) 

H2(g) + 1/2O2(g) → H2O(l) ; ΔfH° = -285.8 kJ/mol …(2) 

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔfH° = +890.3 kJ/mol …(3) 

Here we want one mole of C(graphite) as reactant, so we write down equation (1) as such, we want two moles of H2(g) as reactant, so we multiply equation (2) by 2, we want 

C(graphite) + O2 (g) → CO2(g) ; ΔfH° = -393.5 kJ / mol …(1) 

2H2(g) + O2(g) → 2H2O(l) ; ΔfH° = 2(-285.8 kJ/mol) …(2) 

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ; ΔfH° = +890.3 kJ/mol . … (3) 

Adding we obtain:

C(graphite) + 2H2(g) → CH4(g); ΔfH° = -74.8 kJ/mol


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