Calculate the standard reaction enthalpy for the followingreaction: CH4 (g) + H2

Question

Calculate the standard reaction enthalpy for the followingreaction:
CH4 (g) + H2O (g) --> CO (g) + 3 H2 (g)
Given:
2 H2 (g) + CO (g) --> CH3OH (l) Delta H° = -128.3 kJ/mol
2 CH4 (g) + O2 (g) --> 2 CH3OH (l) Delta H° = -328.1kJ/mol
2 H2 (g) + O2 (g) --> 2 H2O (g) Delta H° = -483.6 kJ/mol
(Please note: the numbers appearing directly after elementalsymbols are subscripts, e.g. CH4 = methane, H2O = water).

Explanation / Answer

Half the secondequation:          CH4+ 1/2O2 ->CH3OH               H = 1/2 (-328.1) = -164.05 KJ/mol Flip the firstequation:               CH3OH -> 2H2 +CO                    H = -(-128.3) = 128.3 KJ/mol Half and flip the third equation: H2O -> H2 +1/2O2                       H = -1/2(-483.6) = 241.8 KJ/mol Final (cross out those that appear on both side): CH4 + H2O -> CO + 3H2 Add up all delta H: -164.05 + 128.3 + 241.8 = 206.05 KJ/mol

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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