Question

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Consider the following reaction at 0 degree C: S(s) + 4NO(g) SO_2(g) + 2N_2O(g) A student places 0.025 moles of solid sulfur and 0.100 moles of nitrogen monoxide gas in a closed 1.00 L container at 0 degree C for several days. Calculate the number of molecules of NO in the 1.00 L container after equilibrium is reached, given the following thermodynamic data:

Answer 1

Given:

T is 0 ⁰ C = 273.15 K

Lets calculate delta G by using delta H and delta S.

Then we calculated equilibrium constant and by using it we can get equilibrium moles of NO.

Delta Hrxn = [ 1 x delta Hf of SO2 + 2 x delta Hf N2O ] – [ 1 x delta Hf of S + 4 x delta Hf NO (g) ]

Lets plug given values in above equation.

Delta Hrxn ={ [ 1 x -297+ 2 x81.6] – [ 1 x 0.00+ 4 x90.4]} kJ

=-495.4 kJ

Delta Hrxn in J

= -495.4 kJ x 1000 J / 1 kJ

= -495400 J

Calculation of delta S

Delta Srxn = [ 1 x delta Sf of SO2 + 2 x delta Sf N2O ] – [ 1 x delta Sf of S + 4 x delta Sf NO (g) ]

Lets plug given values

Delta Srxn = [ 1 x 256+ 2 x220] – [ 1 x 31.9+ 4 x 211]

= -179.9 J / K

lets find delta G

Delta G = Delta H – T delta S

= {- 495400 – ( 273.15 x -   179.9 )} J

= -446260 J

Delta G = -446260 J

Lets use following equation to get K.

Delta G = - RTlnK

here T is in K , R is gas constant , Delta G is free energy change in J

value of R = 8.314 J / (K mol)

Lets plug all the values.

-446260 J = - 8.314 J / K mol x 273.15 K ln K

Lets multiply above equation by -1

446260 J = 8.314 J / K mol x 273.15 K ln K

Now by rearranging we get …

ln K = 446260 J / (8.314 J/ K mol x 273.15)

= 196.5

Lets take exp of both side

K = exp (196.5)

K = 1.0 E 85

This shows that the formation of product is extremely high.

So we calculate limiting reactant.

Mole ration of S to NO is 1 : 4

Moles of NO required for stoichiometric limit is

= 0.025 mol S x 4 mol NO / 1 mol S

= 0.1 mol NO

So both reactant are having exact moles that required for stoichiometric limit.

Since kc value is very high so at there will be any moles of NO remains at equilibrium.