Question

2. 1.487 g of methanol (CH3OH, a liquid at room temperature) is completely combusted at 25 °C in a bomb calorimeter with a heat capacity of 10.22 kJ/°C. The final temperature is found to be 28.291 °C. a. Calculate the standard heat of combustion of CH2OH(D) (ACH ) in kJ/mol at 25 °C. b. Calculate the standard heat of formation of CH3OH() (AFH ) in kJ/mol at 25 °C.

Answer 1

a. The heat of combustion at constant volume can be calculated by using the expression.

Heat liberated by w g of the substance = (- z) (∆T)

∆E = -Z × ∆T × M/W

Where, Z = Heat capacity of calorimeter system (calorimeter +water)

           ∆T = Rise in temperature

           M = Molecular mass of substance

            w = Mass of substance taken

Heat liberated by combustion of the 1 mole of the substance = 10.22*(28.291-25)*32/1.487 = -723.79 KJ/mole

The balanced chemical reaction for the combustion of methanol is

     CH3OH(l) + 3/2 O2(g)      ---------------->     CO2(g) + 2H2O(l), $\Delta$ n=1-3/2 = -1/2

The relation between $\Delta$ H and $\Delta$ U is given by $\Delta$ H = $\Delta$ U + $\Delta$ nRT

                                                                                 = -723.79 + (-1/2)*8.314*10^-3*298

                                                                                 = -725.02 KJ/mole

b. The heat change which occurs when one mole of the substance is formed, under standard conditions, from its constituent elements in their standard states is known as standard enthalpy of formation

    C(s, graphite) + 2H2(g) + 1/2 O2(g) ------------> CH3OH(l)

$\Delta$H^o formation = $\Delta$ Hof CH3OH(l) - $\Delta$ Hof C(graphite) -1/2$\Delta$Hof O2(g)-$\Delta$Hof H2(g)

                         = -238.7 -0-0-0 = -238.7 KJ/mole

The standard enthalpy of formation of CH3OH(l) at 25 oC is -238.7 KJ/mole