Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l)

Question

Explanation / Answer

(B) C(graphite) + O2(g) --> CO2(g) (C) H2(g) + 1/2 O2(g) ---> H2O(l) use the reverse of 1 (A) to get CH3OH as a product use 1 (B) to get 1 mole of graphite as a reactant use 2(C) to get 2 moles of H2 as a reactant -1(A) : CO2(g) + 2 H2O(l) --> CH3OH(l) + 3/2 O2(g) 1 (B) : C(graphite) + O2(g) --> CO2(g 2 (C) : 2 H2(g) + 1 O2(g) ---> 2 H2O(l) note that CO2 --> cancels with --> CO2 3/2 O2 --> cancels with --> 3/2 CO2 2 H2O --> cancels with 2 H2O & leaves you with: C(graphite) + 2 H2(g) + 1/2 O2(g) --> CH3OH(l) so if combining equations "-1(A) & 1 (B) & 2 (C)" gives the equation you wish then combining the energies of "-1(A) & 1 (B) & 2 (C)" gives the energy you wish: - (-726.4) & (-393.5) & 2(-285.8) = -238.7 kJ your answer is dH = -238.7 kJ

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.