The effect of increasing the temperature on the equilibrium constant K for an exothermic reaction can be explained using the Van't Hoff equation:d lnK / d 1/T = -H/Rwhere K is the equilibrium constant, T is the temperature in Kelvin, H is the standard enthalpy change of the reaction, and R is the gas constant 8.314 J/molK .For an exothermic reaction, the enthalpy change H is negative. According to the Van't Hoff equation, if H is negative, then the relationship between lnK and 1/T is positive. This means that as the temperature T increases, the value of 1/T decreases, and the equilibrium constant K decreases as well.Let's consider the formation of a complex ion, such as the formation of hexaaquacopper II ion from copper II sulfate:Cu aq + 6HO l [Cu HO ] aq Assume that the standard enthalpy change H for this reaction is -20 kJ/mol an arbitrary value for illustration purposes . We can use the Van't Hoff equation to calculate the effect of increasing the temperature on the equilibrium constant K .Let's compare the equilibrium constants at two different temperatures: 298 K 25C and 323 K 50C .First, we need to integrate the Van't Hoff equation: d lnK = -H/R d 1/T ln K/K = -H/R 1/T - 1/T Now, we can plug in the values:ln K/K = - -20,000 J/mol / 8.314 J/molK 1/323 K - 1/298 K ln K/K = 1.47K/K = e. 4.34This result indicates that the equilibrium constant K at 50C K is approximately 4.34 times smaller than the equilibrium constant at 25C K . Therefore, increasing the temperature for this exothermic reaction decreases the equilibrium constant K , which means that the formation of the complex ion is less favorable at higher temperatures.