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At room temperature, the equilibrium constant (K) for the formation of a complex ion is 8.0 x 10^-5. If the temperature is increased by 10 degrees Celsius, how will the equilibrium constant be affected? Provide a numerical answer and explanation to support your reasoning.

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To determine how the equilibrium constant  K  will be affected by a temperature change, we need to consider 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 standard enthalpy change of the reaction, R is the gas constant  8.314 J/molK , and T1 and T2 are the initial and final temperatures in Kelvin.However, we do not have enough information to directly calculate the new equilibrium constant  K2  using this equation. We need to know the standard enthalpy change  H  for the formation of the complex ion.Nonetheless, we can provide a general explanation of how the equilibrium constant may be affected by the temperature change:1. If the formation of the complex ion is an endothermic process  H > 0 , increasing the temperature will favor the forward reaction, and the equilibrium constant  K2  will be greater than the initial equilibrium constant  K1 .2. If the formation of the complex ion is an exothermic process  H < 0 , increasing the temperature will favor the reverse reaction, and the equilibrium constant  K2  will be smaller than the initial equilibrium constant  K1 .In summary, without knowing the standard enthalpy change  H  for the formation of the complex ion, we cannot provide a numerical answer for the new equilibrium constant  K2 . However, we can infer that the equilibrium constant will either increase or decrease depending on whether the reaction is endothermic or exothermic, respectively.

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