The combustion of 6 g of ethane released 312 kJ. The combustion of 34.4 g of a mixture of pentane and heptane released 1680 kJ. What is the molar ratio of pentane and heptane in the mixture if it is known that in the homologous series of alkanes the enthalpy of combustion increases by 660 kJ for each mole of CH2 – groups?
We determine the enthalpy of combustion of ethane according to the thermochemical equation:
C2H6 (g) + 7 / 2O2 (g) = 2CO2 (g) + 3H2O (l) + Q
When 6 g of ethane is burned, 312 kJ is released, therefore, when 30 g of ethane is burned, 1560 kJ is released.
We calculate the enthalpy of combustion of pentane and heptane, taking into account that the enthalpy of combustion increases by 660 kJ for one mole of CH2 – groups:
∆Hcomb ^ 0 (C5H12) = 1560 + 660 ∙ 3 = 3540 kJ / mol
∆Hcor ^ 0 (C7H16) = 1560 + 660 ∙ 5 = 4860 kJ / mol
Let’s find the amount of substance of each gas in the mixture.
Let n (C5H12) = x mole, and n (C7H16) = um, then we compose a system of equations:
72x + 100y = 34.4
3540x + 4860y = 1680
x = 0.2 mol, y = 0.2 mol
We find the molar ratio of gases in the mixture:
n (C5H12): n (C7H16) = 0.2: 0.2 = 1: 1
Answer: 1: 1
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