Question

A mixture of 3.00 mol of Cl2 and 3.00 mol of CO is enclosed in a 5.00-L flask at 600.°C. At equilibrium, 3.3% of the Cl2 has been consumed.

CO(g) + Cl2(g) = COCl2(g)

Calculate Kc for the reaction at 600.°C.

Calculate deltaG for the reaction at this temperature. = kJ/mol

CO(g)= deltaH= -110.5 (kJ/mol) deltaG= -137.2 (kj/mol) S= 197.7 (j/molK)

Cl2(g)= S= 223.1 J/molK

COCl2(g)= deltaH= -218.8(kJ/mol) deltaG= -204.6 (kj/mol) S= 197.7 J/molK

Answer 1

Given reaction: co(g)C (g) coc, (g) The equilibrium constant expression for the above reaction is written as the ratio of concentration of products to the reactants. [cocI,] Kc COCI, a) Given number of moles of reactants, Moles of Cl, 3.00 mol Moles of CO=3.00 mol The volume of flask is given as 5.00 L The temperature at which the flask is given as 600°C Calculate the concentrations of given reactants using the given number of moles and volume using the molarity expression number of moles Molarity Volume Concentration of Cl,: number of moles 3.00 mol Molarity C1,] 0.6 M Volume 5.00 L Concentration of CO: 3.00 mol number of moles Molarity Co 0.6 M Volume 5.00 L These are initial concentrations of reactants

To get the equilibrium constant value one needed to calculate the equilibrium concentrations: Get the equilibrium concentrations from ICE table: co(g)C (g) coc, (g) Initial (M Change (M) Equilibrium (M) |0.6-x 0.6 -x 0.6 0.6 0 -X -X We have given that, at equilibrium, 3.3% of Cl, is consumed Therefore, the change is 3.3 % of the given initial concentrations The amount of change in initial concentration will be, 3.3 -x 0.6 M 100 Change in concentration 0.02 M Then the remaining values will be =0.6 x 0.6-0.02 0.58 M leq _ [cola [cocI 0.6 x 0.6-0.02 0.58 M eq = x = 0.02 M leq Now write the equilibrium constant expression and substitute the equilibrium concentrations in the expression as shown below: [COCI, Kc leq 0.02 M (0.58 M) (0.58 M) =0.05945

Therefore, the value of equilibrium constant for the reaction at given temperature is |K =0.05945 b) As we have the equilibrium constant value for the given reaction at 600°C calculate the AG using that value. We don't need any standard values as we don't know at which temperature those values are given we can First calculate AG for the given reaction using the calculated value of K If a reaction involves only gases, the free energy change is related to the equilibrium constant for gases, K2by the following equation: AG= -RT In K, Here, R is universal gas constant and T is temperature. The relation between K, and Kc is, An K, K (RT Here, K is the equilibrium constant in terms of concentration, K, is the equilibrium constant in terms of partial pressures and An is the change in number of moles.

An The moles of gaseous products - the moles of gaseous reactants (1)-(1+1) =-1 Thus, the change in number of moles is -1 K2 K, (RT J (600 C+273 K) = 0.0594x8.3 14 mol K J (873 K) =0.0594x 8.3 14 mol K =8.18x10 Thus, the value of K, is 8.18x10 Calculate the value of AG at 600 C from the following relation AG =-RT ln K T -x873 K x In (8.18x10) mol K -8.314 -4 1 kJ -85020.3 mol 1000 J =85.02 mol J Thus, the value AG of the reaction at 600 °C is 85.02 mol

I have an idea to show how to calculate AG using the standard values. The expression that give AG using the enthalpy and entropy values is AG AH -TAS Here, AH and AS are The standard data given in the question is, change in enthalpy and entropy at standard temperature co (g) AH 110.5 mol AG 137.2 mol J S 197.7 mol K C, (з): AH 0(as Cl2is in its standard form) AG 0(as Clis in its standard form) - J S 197.7 mol K СОСI, (g): AH 218.8 mol J AG -204.6 mol J S 197.7 mol K On referring to above values, the standard values of COCI, (g) seems to have errors So, I have given the calculations using Kc value.