Question

7. -13 points Calculate the standard enthalpy of reaction for the combustion of methane. Round to the nearest whole number. CH (9) + 02 --> CO2(g) + 2H,0(1) kl/mol Compound H, (kJ/mole) CH (9) 75 Co (9) 394 H2O(1) 284 This reaction is: O exothermic O endothermic Submit Answ V iewing Saved Work Revert to Last Response

Answer 1

Standard Enthalpy of a reactions be calculated by using the following relation: At = {t roducte k reactants where t is the standard enthalpy - of the reaction, Eth products is the sum of enthalpies of products and Et reactants is the sum of enthalpy of reactants. Here th in the enthalpy of formation of each species) Now, the given reaction is Сич (4#oy (5) — со (а) деца). Here the products are ca and to. o Ef products (ff) coz + 2(+420 Since there are a moles of the in the reaction so the enthalpy of tion of water has to be multiplied by 2)

of using given formation values of co, enthalpies of and the, we get & # produces (+394.kuJtmoles). + 26-284 flockey = (-394 + 2x.(-284).) k I/mole = (-394 +=563)molen = (-394-5.68.) kJ/mole = - 962 kJ/mole. Now the reactants are Chy and O. The enthalpy of formation of a that is ( 2020 so we get --- Ett rintaner - 7. (Hiety (+4)01. - -= (-75 kJ/mole) +0 kJ/mole / = -75 kJ/mole So standard enthalpy of reaction is - At = Et products - Etf reactants = -962 kj/mole -(-75 kJ/mole) - 962 & Il mole a 75 kHmole (-962 + 75) kj/mole F 887 kJ/mol
So the standard enthalpy of reaction for the combustion of methane is -887 kJ/mol.

When the enthalpy of a reaction is negative , it means energy is released in the reaction which further implies that the reaction is an exothermic reaction and when the enthalpy of a reaction is positive then it means energy is absorbed by the reaction so it is an endothermic reaction.

Since in this reaction the net enthalpy of the reaction comes out to be a negative number (-887) , so the reaction is exothermic.