Question

Consider the reaction ch3oh(l)→ch4(g)+1/2o2(g). part a calculate δrh∘298.

Answer 1

`\Delta _rH^o=163.53kJ/mol`

Step-by-step explanation:

Hello,

In this case, the proper way to compute the enthalpy of reaction is as shown below:

`\Delta _rH^o=\Delta _fH^o_(CH_4)+(1)/(2) \Delta _fH^o_(O_2)-\Delta _fH^o_(CH_3OH)`

Wherein all the enthalpy changes are referred to the formation enthalpy which are subsequently replaced for each compound at its corresponding phase as follows:

`\Delta _rH^o=-74.87kJ/mol+0kJ/mol-(-238.4kJ/mol)\\\Delta _rH^o=163.53kJ/mol`

Best regards.

Answer 2

To determine the standard heat of reaction, ΔHrxn°, let's apply the Hess' Law.

ΔHrxn° = ∑(ν×ΔHf° of products) - ∑(ν×ΔHf° of reactants)
where
ν si the stoichiometric coefficient of the substances in the reaction
ΔHf° is the standard heat of formation

The ΔHf° for the substances are the following:
CH₃OH(l) = -238.4 kJ/mol
CH₄(g) = -74.7 kJ/mol
O₂(g) = 0 kJ/mol

ΔHrxn° = (1 mol×-74.7 kJ/mol) - ∑(1 mol×-238.4 kJ/mol)
ΔHrxn° = +163.7 kJ