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

Question

A scientist measures the standard enthalpy change for the following reaction to be 201.6 kJ:

CH₄(g) + H₂O(g)3H₂(g) + CO(g)

Answer 1

Previously answered. If you don't understand this, or have a problem with it, it would be helpful if you state your issue so that we might be more helpful. The standard enthalpy of formation of the reaction is equal to the standard enthalpies of formation of the products minus those of the reactants, as described below.

You should look up the ∆Hº_f for H₂(g) which will be zero, and also for CH₄(g) and H₂O(g). Then use the fact that ∆Hºrxn = ∑∆Hº_f products - ∑∆Hº_f reactants

201.6 kJ = ∆Hº_f CO(g) - [∆Hº_fCH4(g) + ∆Hº_fH₂O(g)] and solve for ∆Hº_f CO(g)