show your work A certain substance has a heat of vaporization of 67.49 kJ/mol. A

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A certain substance has a heat of vaporization of 67.49 kJ/mol. At what Kelvin temperature will the vapor pressure be 3.00 times higher than it was at 327 K? At 25 degree C, the equilibrium partial pressures for the following reaction were found to be PA = 4.80 atm, PB = 2.90 atm, PC = 2.90 atm, and PD = 9.10 atm. A (g) + 2B (g) rightarrow C (g) + D (g) What is the standard change in Gibbs free energy of this reaction at 25 degree C. What is the value of K for this aqueous reaction at 298 K? Delta G degree = 24.05 k J/mol Under standard conditions (298 K and 1 atm), which statement is true? diamond converts to graphite spontaneously graphite converts to diamond spontaneously none of the above How can the spontaneity of the reaction be reversed? increase the temperature decrease the temperature none of the above Calculate the standard change in Gibbs free energy for the following reaction at 25 degree C. C2 H2 (g) + 4Cl2 (g) rightarrow 2C Cl4 (l) + H2 (g) For a gaseous reaction, standard are 298 K and a partial pressure of 1 atm for all species. For the reaction the standard change in Gibbs free energy is Delta G degree = -72.6 k J/mol. What is Delta G for this reaction at 298 K when the partial pressures are P c2H6 = 0.300 atm, P H2 = 0.250 atm, and P CH4 = 0.900 atm

Explanation / Answer

Q1) Use the Clausius-Clapeyron equation

ln(p1/p2)=Hvap/8.3145 (1/T1-1/T2)

ln(3) = (67490/8.3145)(1/327 - 1/T2)

T2 = 342.14 K

Q2) The appropriate equation to begin with is:

Delta G = Delta Go + R T ln Q

where Q has the form of the equilibrium constant.

Now, at equilibrium, DeltaG = 0, so

Delta Go = - RT ln Keq

For this reaction, Kp = (PD)(PC) / (PA)(PB)^2

Kp = ((9.10)(2.90))/((4.80)(2.90)^2) = 0.65

Now, Delta Go = - RT ln Kp = -8.314 J/molK X 298K X ln 0.65 = -1067 J/mol = -1.067 kJ/mol

Q3) Delta Go = - RT ln K

Delta Go / -RT = ln K

K = e^(Delta Go/-RT)

K = e^(24050/-((8.314)(298)))

K = 6.09 x 10^-5

Q4) diamond converts to graphite spontaneously because Delta H to be -1.9 kJ/mol
Delta S to be 3.3 J/mol k
and Delta G to be -2.9 kJ/mol

Q5) C

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