To understand how free energy relates to equilibrium under standard and nonstand

Question

To understand how free energy relates to equilibrium under standard and nonstandard conditions.

The drive toward equilibrium is what makes a reaction spontaneous. This drive is quantified by ?G, a state function known as free energy. A negative value of ?G indicates a spontaneous reaction. Standard free energies of formation, ?G?f, can be used to calculate the standard free energy of reaction, ?G?rxn, for any given chemical equation.

Standard free energy is related to the equilibrium constant K by the equation ?G?=?RTlnK, where T is the Kelvin temperature and R is the gas constant equal to 8.314 J/(mol?K). Standard thermodynamic conditions are 1 atm and 298 K.

Free energy is related to the reaction quotient Q by the equation ?G=?G?+RTlnQ, where T is the Kelvin temperature and R is the gas constant equal to 8.314 J/(mol?K).

Nitrosyl chloride formation

Chlorine gas, Cl2(g), reacts with nitric oxide, NO(g), to form nitrosyl chloride, NOCl(g), via the reaction

Cl2(g)+2NO(g)?2NOCl(g)

The thermodynamic data for the reactants and products in the reaction are given in the following table:

The diagram below serves as a pictorial depiction of the relationship between the free energy and standard free energy in the equation ?G=?G?+RTlnQ. Point A is at standard conditions, which means all reagents have a partial pressure of 1 atm, as shown in the equation below point A. Points B and C are at nonstandard conditions. Point X indicates the point at which equilibrium lies. The axis labeled Q refers to the reaction quotient. The actual value of Q at the point of equilibrium is Kp, which is the equilibrium constant at 298 K.

Part A

Using standard free energy of formation values given in the introduction, calculate the equilibrium constant Kp of the reaction

Cl2(g)+2NO(g)?2NOCl(g)

The standard free energy of the reaction represents the drive the reaction has under standard conditions to move toward equilibrium from point A to point X in the diagram.

Express the equilibrium constant numerically using three significant figures.

Explanation / Answer

delta Go = Goproducts - Go reactants

= 2* 66.3 -2 * 86.71 = -40.82kJ/mol= -40.82E3 J/mol

Delta Go = -RTlnKp = -40.82E3 J/mol

R = 8.314 J/K/mol T = 298K as standard conditions

So ln Kp =-( -40.82E3/(8.314*298)) = 16.475 J/mol

Kp = 14300538.71 = 1.43E7

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