Question

The change in enthalpy for a reaction is -25.8kJ. The equilibrium constant for the reaction is 1400 at 25 degrees Celsius. What is the equilibrium constant at 382 degrees Celsius?

Answer 1

According to Van't Hoff equation ,

ln (K' / K ) = -(H / R) x [(1/T')-(1/T)]

Where

K = Equilibrium constant at temperature T = 1400

K' = Equilibrium constant at temperature T' = ?

T = initial temperature = 25^oC = 25+273 = 298 K

T' = final temperature = 382 ^oC = 382+273 = 655 K

H = change in enthalpy for the reaction = -25.8kJ = -25.8x1000 J          Since 1kJ = 1000 J

                                                            = -25800 J

R = gas constant = 8.314 J/(mol-K)

Plug the values we get

ln (K' / 1400) = -(-25800 / 8.314 ) x [(1/655)-(1/298)]

                   = -5.67

   K' / 1400 = e ^-5.67

                = 3.43x10^-3

             K' = 1400 x 3.43x10^-3

                = 4.80

Therefore the equilibrium constant at 382 degrees Celsius is 4.80