The thermochemical equation for the combustion of butane is givenn below. C₄H₁₀(g) + 13/2 O₂(g) ? 4 CO₂(g) + 5 H₂O( $\ell$ )
?_rH° = -2877 kJ/mol-rxn
What is the enthalpy change for the reaction below?
8 CO₂(g) + 10 H₂O( $\ell$ ) ? 2 C₄H₁₀(g) + 13 O₂(g)

Question

The thermochemical equation for the combustion of butane is givenn below. C₄H₁₀(g) + 13/2 O₂(g) ? 4 CO₂(g) + 5 H₂O(

\ell

)
?_rH° = -2877 kJ/mol-rxn
What is the enthalpy change for the reaction below?
8 CO₂(g) + 10 H₂O(

\ell

) ? 2 C₄H₁₀(g) + 13 O₂(g)

Quizplus · Accepted Answer

The given reaction is the reverse of the combustion of butane, which means the sign of the enthalpy change is reversed. Since the combustion of one mole of butane releases 2877 kJ, the reverse reaction for two moles of butane will absorb 2 x 2877 kJ = 5754 kJ.

The Thermochemical Equation for the Combustion of Butane Is Givenn ℓ\ellℓ

The Thermochemical Equation for the Combustion of Butane Is Givenn $\ell$