i need help with this question please Gold reacts with oxygen according to the f

Question

i need help with this question please

Gold reacts with oxygen according to the following reaction: At 25 degree C, the Gibbs Free Energy Change (delta G) for the reaction is +78.7 kJ kmol^-1. What factors affect the equilibrium yield of a chemical reaction? Estimate the value of oxygen pressure required oxidizing gold. Is the reaction spontaneous in air? Estimate the temperature required for gold oxidation if the Enthalpy Change (delta H) for the reaction at 25 degree C is -13.0 kJ kmol^-1. Specify any assumptions you made for calculations.

Explanation / Answer

Given reaction is 2Au(solid) + 3/2 O2 (g) === > Au2O3 solid

Gibbs free energy = +78.7 kJ/mol

a)

Given reaction has pressure to reactants side and rest species are solids.

So, we use kp

Kp = 1 / p(O2) ^ (3/2)

Let’s also write Qp expression,

Qp = 1 / p(O2) ^ (3/2)

For this reaction to occur Qp value should be less than kp and that is done only by increasing pressure of O2 gas.

So, one of the factors responsible here is Pressure of O2 gas.

Let’s also show the delta G of reaction.

Delta G = delta G ( std ) + RTlnQp

Here R is gas constant. Value is 8.314 J / K mol

If the pressure of O2 will be higher then Qp will be less than 1 and then the value of Detla G will goes towards negative which will help to become reaction spontaneous.

Temperature: As we can see that Temperature is involved in second term so when we lower the temperature second terms will be less and so it will also help to decrease free energy of reaction.

Answer: Important factor that drive reaction would be , Temperature (lowering) and pressure O2( increasing)

b)

lets calculate the value of pressure required to become reaction spontaneous.

Reaction becomes spontaneous when delta G just start to becomes less than zero. So we substitute delta G = 0 ,

0 = delta G (std ) + (8.314 J / K mol x 298.15 K x ln Qp

Detal G = - 2478.819 x ln Qp

78700 = - 2478.819 x ln Qp     ( delta G(std) used in J )

Qp = 1.69 E-14

Since Qp = 1/ p(O2) ^ 3/2

P (O2) = 1555517603.8 Pa

Pressure in atm

1555517603.8 Pa x 9.87 E-6 atm / 1 Pa

= 1.54 x 10^4 atm

So, pressure of O2 required = 1.54 E4 atm

This pressure is far high from air pressure so it cant oxidize gold in air.

c)

If Delta H = -13.0 kJ /mol

Delta H and delta G do not change over Temperature ( approximation )

Detla G = delta H -Tdelta S

Lets first calculate delta G

Delta G = delta G ( std)- RTlnQp

Delta G = 78700 + 8.314 x 298.15 K x ln ( 1.69 E-14 )

93.0 J

Les calculate delta S

Delta S = delta H / T = -13000 J /mol / 298.15 K = 43.6 J / K mol

Lets plug in all the values to get T

93.0 = -13000 – T x ( 43.6 )

T = -300.298 K

So temperature = -300.3 K

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