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What is the change in equilibrium constant for the formation of a complex ion when the temperature is increased from 25°C to 50°C, given that the enthalpy of reaction is -20 kJ/mol and the entropy change is 80 J/(mol·K)?

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To calculate the change in the equilibrium constant  K  for the formation of a complex ion when the temperature is increased from 25C to 50C, we can use the van't Hoff equation:ln K2/K1  = -H/R *  1/T2 - 1/T1 where:- K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively- H is the enthalpy change of the reaction  in J/mol - R is the gas constant  8.314 J/ molK  - T1 and T2 are the initial and final temperatures in Kelvin  K First, convert the temperatures from Celsius to Kelvin:T1 = 25C + 273.15 = 298.15 KT2 = 50C + 273.15 = 323.15 KNext, convert the enthalpy change from kJ/mol to J/mol:H = -20 kJ/mol * 1000 J/kJ = -20000 J/molNow, plug the values into the van't Hoff equation:ln K2/K1  = - -20000 J/mol  /  8.314 J/ molK   *  1/323.15 K - 1/298.15 K ln K2/K1   10.12To find the change in the equilibrium constant, we need to find the ratio K2/K1:K2/K1 = e^10.12   24800Since we are looking for the change in the equilibrium constant, we can say that the equilibrium constant has increased by a factor of approximately 24800 when the temperature is increased from 25C to 50C.

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