Given the following thermodynamic data for the reaction at 25C: 2 C2H6 (g) + 7 O

Question

Given the following thermodynamic data for the reaction at 25C:

                                       2 C2H6 (g) + 7 O2 (g) ----> 4 CO2 (g) + 6 H2O (g)

(Delta)fH0/kJ/mol         -165.2                                     -393.5          -241.8

S0 / J/(mol K)                   229.5           205.0               213.6            188.7

a.) Find the value of (delta)H0 for the reaction at 25C.

b.)  Find the value of (delta)S0 for the reaction at 25C.

c.) Find the value of (delta)G0 for the reaction at 25C.

d.) Find the value of (delta)G of the reaction at 25C if the pressure of CO2 and that of H2O gases are 1 atm each , but the pressure of C2H6 and that of O2 gases are each 0.001 torr. Hint: remember that the activity of a gas numerically equals its partial pressure in atmospheres.

e.) Will the reaction under the conditions in part d.) go forward , backward or will it be at equilibrium under conditions given in part d.) Explain.

Please show work I don't understand how it is done, and I need to study for my final

Explanation / Answer

*** for calculating delta Ho and Delta So u just need to subtract the products - reactants they beig multiplied by their resective number of moles.

2 C2H6 (g) + 7 O2 (g) ----> 4 CO2 (g) + 6 H2O (g)

(Delta)fH0/kJ/mol...... -165.2 0 -393.5 -241.8

S0 / J/(mol K) ..........229.5 205.0 213.6 188.7

a) Delta Ho = ( 6 * -241.8 + 4 * -393.5 ) - (2 * -165.2) - 7 * 0

= (-2694) KJ/mol

........for oxygen since it is in elemental form heat of formation is ZERO.

b) delta So = 6 * 188.7 + 4 * 213.6 -(7 * 205 + 2 * 229.5)

= (92.6)J/(mol K)

=(0.0926 )kJ/(mol K)

C)Delta Go = Delta Ho - T delta so

=(-2694) - 298 * (0.0926)

= - 2721 KJ / mol

D ) We can now express the G

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