Question

7. Consider the following data: C2H6(g) -84.7 32.9 H2O(l) 285.9 -237.2 CO2(g) -393.5 -394.4 Ho kl/mole) a) Calculate the entropy change for the combustion of 1.0 mole of C,H,(g) at 25°C. b) Are the products more or less ordered than the reactants at 25 c) Is the reaction spontaneous at 25°C? d) Wh at is the maximum amount of useful work which can be obtained from the process at 25°C? e) If the process is carried out irreversibly at 25°C, under constant volume f) g) What is the heat flow if the process is carried out reversibly at 25°C? what is the heat flow? If the what is the heat flow? process is carried out irreversibly under constant pressure at 25°C h) If the process occurs under constant pressure at 25°C, what is the PAV work?

do part f

Answer 1

Answer f:

Heat flow at constant temperature for a reaction (enthalpy of combustion in our case) is called enthalpy of reaction ($\Delta$H_rxn )

and

$\Delta$H_rxn = sum of enthalpy of formation of products - sum of enthalpy of formation of reactants      ...(1)

here the reaction for combustion is,

C₂H₆    +    7/2 O₂      ---->     2 CO₂     +        3 H₂O

Since enthalpy of formation data is given

So, following (1) we get

$\Delta$H_rxn = 2 x $\Delta$H^o_f( CO₂)   + 3 x $\Delta$H^o_f( H₂O)     -     1 x $\Delta$H^o_f( C₂H₆)

putting the values from above table will give

$\Delta$H_rxn = 2 x (-393.5 kJ/mol) +   3 x (-285.9 kJ/mol) - 1 x (-84.7 kJ/mol) = (-787.0 - 857.7 + 84.7) kJ/mol = -1560.0 kJ/mol

so heat change at 25^oC for burning 1 mole of C₂H₆ at constant pressure will be -1560.0 kJ/mol ( negative sign means, heat will be released.