infr.uni For the system, 2 NO2(g)N204(9), AG 5.40 kJ/mol. If the system was init

Question

infr.uni For the system, 2 NO2(g)N204(9), AG 5.40 kJ/mol. If the system was initialized using enough NO2 to create a partial pressure of 0.500 atm and enough N204 to create a partial pressure of 0.350 atm, calculate the value of 6(T-298K) for these non-standard conditions prevailing at the start. k)/mol Use correct number of significant digits, the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work , you Question Attempts: 0 of 3 used SAVE FOR LATER Earn Maximum Points available only If you answer this question correctly in three attempts or less.

Explanation / Answer

Ans. # Given, pNO2 = 0.500 atm                ; pN2O4 = 0.350 atm

Now,

            Equilibrium constant, Keq = pN2O4 / (pNO2)2

                                    Or, Keq = 0.350 / (0.500)2

                                    Hence, Keq = 1.400

# Using, dG = dG0’ + RT (ln Keq)                         - equation 1

Where, dG = free energy change under experimental (non-standard) conditions

dG0’ = standard free energy change of the reaction

R = universal gas constant = 0.001987 kcal mol-1K-1 = (0.0083146 kJ mol-1 K-1)

T = temperature in kelvin

Keq = equilibrium constant

Putting the values in equation 1-

            dG = -5.40 kJ mol1 + (0.0083146 kJ mol-1 K-1) x 298 K x ln 1.40

            Or, dG = -5.40 kJ mol1 + 0.834 kJ mol-1

            Hence, dG = -4.566 kJ mol-1

Therefore, free energy change for the reaction under given conditions = -4.566 kJ mol-1

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