Calculate the equilibrium constant for the following reaction: co2+ (aq) + zn(s> co (s) + zn2+ (aq)

The of the reaction is

Explanation:

For the given half reactions:

Oxidation half reaction:

Reduction half reaction:

Net reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

and,

Equating these two equations, we get:

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

= standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction =

= equilibrium constant of the reaction = ?

Putting values in above equation, we get:

Hence, the of the reaction is

it is a solid

explanation:

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole