Question

Calculate the enthalpy of the formation of butane, C4H10, using the balanced chemical equation and the standard value below:

4C(s) + 5H2(g) => C4H10(g)

Standard enthalpy of formation values:

(Delta Triangle)H^0 of C(s)= -393.5kJ/mol

(Delta triangle)H^0f of H2(g)=-285.8 kJ/mol

(Delta triangle)H^0f of C4H10(g)=-2877.6kJ/mol

Answer 1

Answer:

+125.4 KJmol-1

Explanation:

∆H C4H10(g) = -2877.6kJ/mol

∆H C(s)=-393.5kJ/mol

∆H H2(g) = -285.8

∆H reaction= ∆Hproducts - ∆H reactants

∆H reaction= (-2877.6kJ/mol) - [4(-393.5kJ/mol) +5(-285.8)]

∆H reaction= +125.4 KJmol-1