Question

calculate the standard enthalpy of combustion of methane,given that standard enthalpy of carbon and hydrogen are 393.5 kj mol and -283.83kj mol.standard enthalpy of formation of methane is 75.16 kj mol.?

Answer 1

Answer on Question #71306 – Chemistry – Other

Task:

Calculate the standard enthalpy of combustion of methane. Given that standard enthalpy of carbon and hydrogen are -393.5 kJ/mol and -283.83 kJ/mol. Standard enthalpy of formation of methane, is -75.16 kJ/mol?

Solution:

Write the equation and balance the reaction of combustion of methane:

$$CH_4 + 2O_2 = CO_2 + 2H_2O$$

Write the equation and balance the reaction of formation of methane:

$$C + 2H_2 = CH_4, \quad \Delta_f H(1) = -75.16\,kJ/mol.$$

Write the equation and balance the reaction of combustion of carbon:

$$C + O_2 = CO_2, \quad \Delta_f H(2) = -393.5\,kJ/mol.$$

Write the equation and balance the reaction of combustion of hydrogen:

$$H_2 + \frac{1}{2}O_2 = H_2O, \quad \Delta_f H(3) = -283.83\,kJ/mol.$$

The calculation for the heat of combustion for methane using Hess's law:

$$\text{Heat of reaction} = \Delta_c H = -\Delta_f H(1) + \Delta_f H(2) + 2*\Delta_f H(3)$$ $$\text{Heat of combustion} = \Delta_c H(CH_4) = 75.16 - 393.5 + 2*(-283.83) = -886\,kJ/mol$$

So, the heat of combustion, that is, the heat of reaction, is -886 kJ per mole of methane.

Answer: The standard enthalpy of combustion of methane is -886 kJ/mol.

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