22. The table shows standard enthalpy changes of combustion, Delta_(c) H . Substance & Delta_(mathrm(c)) mathrm(H) / mathrm(kJ) mathrm(mol)^-1 mathrm(C)(mathrm(s)) & -393.5 mathrm(H)_(2)(mathrm(~g)) & -285.8 mathrm(C)_(4) mathrm(H)_(10)(mathrm(~g)) & -2876.5 What is the enthalpy change for the following reaction? A 4 mathrm(C)(mathrm(s))+5 mathrm(H)_(2)(mathrm(~g)) arrow mathrm(C)_(4) mathrm(H)_(10)(mathrm(~g)) A -2197.2 mathrm(~kJ) mathrm(~mol)^-1 B -126.5 mathrm(~kJ) mathrm(~mol)^-1 C +126.5 mathrm(~kJ) mathrm(~mol)^-1 D +2197.2 mathrm(~kJ) mathrm(~mol)^-1

The problem is asking us to find out the enthalpy change for the following reaction: We try to explain each step while solving this problem:According to the concept of Hess's Law, the enthalpy change for this reaction can be found by subtracting the total enthalpy change of the reactants from that of the products. For this instance, we multiply and sum the standard enthalpy changes of the reactants to get their total enthalpy changes. Reaction enthalpy is equal to the sum product of the amount of the substance involved and their standard enthalpies:For the reactants we have:For C(s), there are 4 moles, each with a -393.5 kJ/mol heat of combustion, getting .For H2(g), there are 5 moles, each with a -285.8 kJ/mol heat of combustion, getting .Summing the two together we get: The reactant enthalpy change is .And then,For the products, we only have C4H10(g), which is given as -2876.5 kJ/mol, so this is the total enthalpy change for the reaction products.So the total enthalpy change for the reaction ∆H is given as ∆H = (sum of enthalpies of products) - (sum of enthalpies of reactants).Substituting our results into this formula, we have:∆H= (-2876.5) - (-3003.0 ) = +126.5 kJTherefore, this implies that 126.5 kJ of heat is absorbed by the reaction signifying it is an endothermic reaction. For this reaction, the enthalpy change would be +126.5 kJ/mol.So the 【Answer】is C , which aligns with the experts' answer.