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

Question

Chemistry

Twinkie

30 June, 10:44

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

+2

Answers (1)

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Cheyanne Daugherty · Answer 1

+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