Question

A chemistry student is given the following thermochemical equations: Equation 1: H2 (g) + 1/2O2 (g) → H2O(l) ΔH1 = -285.8 kJ/mol Equation 2: C(s) + O2 (g) → CO2 (g) ΔH2 = -393.5 kJ/mol Equation 3: 2H2 (g) + O2 (g) → 2H2O(l) ΔH3 = -571.6 kJ/mol Using Hess's Law, calculate the enthalpy change for the following reaction: C(s) + 2H2 (g) → CH4 (g)

Answer 1

To solve this problem, we need to manipulate the given thermochemical equations in such a way that when they are added together, they will give us the desired reaction: C s  + 2H2  g   CH4  g .First, let's reverse Equation 1 to get the formation of H2  g  and 1/2O2  g  from H2O l :H2O l   H2  g  + 1/2O2  g  H1' = +285.8 kJ/molNow, let's multiply Equation 1' by 2 to get the formation of 2H2  g  and O2  g  from 2H2O l :2H2O l   2H2  g  + O2  g  H1'' = +571.6 kJ/molNext, we need to form CH4  g  from C s  and 2H2  g . We can do this by adding Equation 1'' and Equation 2:C s  + O2  g   CO2  g  H2 = -393.5 kJ/mol2H2O l   2H2  g  + O2  g  H1'' = +571.6 kJ/mol----------------------------------------------C s  + 2H2  g  + 2H2O l   CO2  g  + 2H2  g  + O2  g Now, let's subtract Equation 3 from the above equation to eliminate the 2H2O l  term:C s  + 2H2  g  + 2H2O l   CO2  g  + 2H2  g  + O2  g - [2H2  g  + O2  g   2H2O l  H3 = -571.6 kJ/mol]----------------------------------------------C s  + 2H2  g   CH4  g Now, we can calculate the enthalpy change for the desired reaction:H = H1'' + H2 - H3H = +571.6 kJ/mol - 393.5 kJ/mol -  -571.6 kJ/mol H = +178.1 kJ/molSo, the enthalpy change for the reaction C s  + 2H2  g   CH4  g  is +178.1 kJ/mol.