Question

What is the enthalpy change for the reaction 2CuO(s) + C(s) → 2Cu(s) + CO2(g) if the enthalpy of formation for CuO(s) is -159.2 kJ/mol, the enthalpy of formation for CO2(g) is -393.5 kJ/mol, and the standard enthalpy of formation for Cu(s) is 0?

Answer 1

To calculate the enthalpy change for the reaction, we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps in the reaction. In this case, we can use the enthalpy of formation values given for the reactants and products.The reaction is: 2CuO s  + C s   2Cu s  + CO2 g First, we need to find the total enthalpy change for the reactants and products.For the reactants:2 moles of CuO s : 2 *  -159.2 kJ/mol  = -318.4 kJ1 mole of C s : Since it is in its standard state, its enthalpy of formation is 0.For the products:2 moles of Cu s : Since it is in its standard state, its enthalpy of formation is 0.1 mole of CO2 g : -393.5 kJ/molNow, we can calculate the enthalpy change for the reaction:H =   enthalpy of products  -   enthalpy of reactants H =  -393.5 kJ  -  -318.4 kJ H = 75.1 kJThe enthalpy change for the reaction is 75.1 kJ.