Chemistry question

Question

A scientist measures the standard enthalpy change for the following reaction to be 215.0 kJ :

CH₄(g) + H₂O(g)3H₂(g) + CO(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H₂O(g) is ____________ kJ/mol.

Answer 1

The reaction involves breaking up the reactants and forming the products. If you have the values for the enthalpies (or bond energies) of the reactants and products.

Enthalpy of reaction = energy needed to break bonds in products - energy needed to break bonds in products

OR heat of formation of products - heat of formation of reactants.

Answer 2

The standard enthalpy of the reaction is the sum of the enthalpies of formation for the products - the sum of the enthalpies of formation of the reactants multiplied by the coefficients in the balanced equation.

3 x 0 + 1 x ΔH_{f CO} -( ΔH_{f CH4} + ΔH_{f H2O}) = 215.0 kj

From the internet I found ΔH_{f CH4} = -74.8 kj/mole, ΔH_{f CO} = -110.5 kj/mole)

Solve for ΔH_{f H2O} = -(215.0 kj - 0 + 110.5 - 74.8) = -250.7 kj/mole

The result found in my reference was -241.8 kj/mole

I am not sure why there is a discrepancy.

Answer 3

∆H^o_frxn=∑H^o_f [Products]- ∑H^o_f [Reactans]

We know

∆H^o_f of H₂(g)=0kJ/mol

∆H^o_f of CO(g)=-110.5kJ/mol

∆H^o_f of CH₄(g)=-74.8kJ/mol

Substitute the values in the Hess’s law equation

215kJ=[(3*0)+(1*-110.5kj)]-[(1*-74.8kj)+(1*H_f^o of H_2 O(g))]

(215+110.5-74.8)kJ=- ∑H^o_f[Reactans]

∆H_f^o (H_2 (g))=-250.7kJ.mol^(-1)