Question

Consider the following gas-state dimerization at 300k where M is some molecule. You have observed the electronic absorption spectra are efficiently distinctly so that the partial pressures.

please write a detailed solution to Part A and Part B.

1. Consider the following gas-state dimerization reaction at 300 K: 2 M(g) M2(g) where M is some molecule. You have observed that the electronic absorption spectra of M(g) and Mz(g) are sufficiently distinct so that the partial pressures of M and M2 can be determined using a UV-Vis absorption spectrometer. By constructing proper calibration curves, you determine that the partial pressure of M(g) in atm, pM, is related to the absorption at 450 nm, A450, via A450-арм. Similarly, you find the partial pressure of M2(g) in atm, pMa, ls related to the absorption at 650 nm, Asso, via A6so -bpM.The constants a and b are found to be 4 atm1 and 2 atm, respectively a) You measure the UV-Vis spectrum of the equilibrium mixture and find that 300 K, A450 -0.01 and A6so -0.2, giving a reddish color. Calculate the standard reaction Gibbs free energy at 300 K b) You raise the temperature to 325 K and find that the gas mixture has turned yellowish, yielding a spectrum with A450 -0.3 and A6so 0.01. Calculate the standard reaction entropy at 300 K. Any assumptions you make should be stated explicitly.

Answer 1

a) As given in the example, partial pressure of M₂ and M can be found at 300K. a and b values are already given in the example. These are constant values and hence, will not change for the particular wavelength.

Now, for molecule M, Absorbance and partial functions are related as A₄₅₀ = aP_M ( where a = 4 atm^-1)

From the given value of A₄₅₀ in the question, 0.01 = 4 $\times$ P_M

P_M = 0.01/4 = 0.0025 atm

Similarly, for M₂ molecule, A₆₅₀ = bP_^M2 ( where b = 2 atm^-1)

From the given question, 0.2 = 2 $\times$ P_^M2

P_^M2 = 0.2 / 2 = 0.1 atm

Now, at equilibrium, the equilibrium constant can be written in terms of partial pressure as follows:

2M (g) $\rightleftharpoons$ M₂ (g)

Kp = [P_^M2] / [P_^M]²

= 0.1 / ( 0.0025)²

= 16,000

Now, at equilibrium, the standard reaction Gibbs free energy, $\Delta$G^o = - RT ln Kp ( R is gas constant and T is temperature)

  $\Delta$G^o = - RT ln Kp

= - (8.314 JK^-1) $\times$ 300 K $\times$ ln (16,000)

= - 24.144 kJ/mol

b) At 350 K, for molecule M, Absorbance and partial functions are related as A₄₅₀ = aP_M ( where a = 4 atm^-1)

From the given value of A₄₅₀ in the question, 0.3 = 4 $\times$ P_M

P_M = 0.3/4 = 0.075 atm

Similarly, for M₂ molecule, A₆₅₀ = bP_^M2 ( where b = 2 atm^-1)

From the given question, 0.01 = 2 $\times$ P_^M2

P_^M2 = 0.01 / 2 = 0.005 atm

Kp = [P_^M2] / [P_^M]²

= 0.005 / ( 0.075)²

= 0.89

Now, $\Delta$G^o = - RT ln Kp

= - (8.314 JK^-1) $\times$ 300 K $\times$ ln (0.89)

= 290.51 J/mol

Assume that there is neither heat consumption or heat emission during the process i.e. $\Delta$H^o = 0

$\Delta$G^o = $\Delta$H^o - T $\Delta$S^o

$\Delta$G^o = 0 - T $\Delta$S^o

$\therefore$$\Delta$S^o_{325 K} = -  $\Delta$G^o / T = - 290.51 / 325 = - 0.894 J mol^-1 K^-1

  $\Delta$S^o_{300 K} = -  $\Delta$G^o / T = 24.144 / 300 = 0.08 kJ mol^-1 K^-1