How do you calculate the enthalpy of formation of butaine

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Answer: "Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical equation and tables displaying information about enthalpy of formation/combustion. Write out the solution according to Hess's law." That's all I'm given. The key says that the answer is -125.4 kJ. C4H10 + 6.5 O2 = 4CO2 + 5 H20 The heat of combustion for one mole of butane is -2877 K. Heats of formation from table C02 = -393.5 Kj/mole, H20 (g) -241.8 Kj/mole. Heat of combustion of butane = sum heats of formation products minus heats of formation reactants -2877 Kj = 4(-393.5Kj) + 5(-241.8Kj) minus heat or enthalpy of formation butane.

The enthalpy formation oxygen an element is assigned a value of zero.

-2877 Kj = -1582 Kj + -1209 Kj minus X

-2877 = - 2791 Kj minus X

X = -86 Kj. The heats of formation for various compounds vary a little from table to table, so this does not quite agree with your answer.

This is just an example

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Considering the Dalton's law, the total pressure of the tank is 5.95 atm.

<h3>Dalton's law</h3>

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So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

$P_{T} =P_{1} + P_{2} + ... + P_{n}$