Question

# Calculate the heat of reaction in kcal at 25 °C and 1 atm for the following...

Calculate the heat of reaction in kcal at 25 °C and 1 atm for the following reaction per mol from the given data in the table below. C3H8 (l) + O2 (g) à CO2 (g) + H2O (l) Compound DHo f (kcal/mol) DHv at 25 °C (kcal/mol) C3H8 (g) 24.820 3.823 CO2 (g) 94.052 1.263 H2O (g) 57.798 10.519 Hint: pay attention to phases!

The standard heat of reaction is given by the following equation:

The balanced chemical equation is given by:

The governing equation gets reduced to:

Given that:

ΔHH2O(l)o = ΔHH2O(g)o + ΔHH2Ov = - 57.798 + (- 10.519) = - 68.317 kcal/mol

ΔHCO2(g)o​​​​​​​ = - 94.052 kcal/mol

ΔHC3H8(l)o​​​​​​​ = ΔHC3H8(g)o​​​​​​​ + ΔHC3H8v​​​​​​​ = - 24.82 + (- 3.823) = - 28.643 kcal/mol

ΔHO2(g)o​​​​​​​ = 0 (By definition)

Sustituting these values in the governing equation.

ΔHoreaction = (3 X (- 94.052) + 4 X (- 68.317) - (-28.643) - 5 X 0)

ΔHoreaction = - 282.156 - 273.268 + 28.643

ΔHoreaction = -526.781 kcal / mol

which is the required standard heat of reaction.

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