Question

Calculate the heat energy energy released in J/g if the standard enthalpy of combution of butane is 2658KJ/mol according to following combustion reaction C4H10+13/2O2-- 4CO2+5H2O

Delta cH*= -2658.0 KJ/mol

Answer 1

Reaction Involved-
C­₄H₁₀ + $\frac{13}{2}$O­₂ → 4CO₂ + 5H₂O

Given-
∆H = -2658.0 kJ/mol

Enthalpy of combustion of butane,
∆H = -2658 kJ/mol = -2658 $\times$ 10 ³ J/mol

Consider molar mass of butane to find enthalpy in J/g
Molar mass of butane = 12$\times$4 + 10$\times$1 = 58 g/mol
​This means 58 g butane on combustion would give energy = -2658 $\times$10 ³ J
Thus 1 g butane on combustion would give energy = $\frac{-2658\times {10}^{3}J}{58}$ = 45.83$\times$10 ³ J / g
Thus Enthalpy of combustion of butane in J/g
∆H = 45.83 x 10³ J/g