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Calculate the equilibrium constant for the electrochemical reaction: Zn(s) + 2H^+(aq) → Zn^2+(aq) + H2(g) when the concentration of Zn^2+ is 0.25 M, [H^+] is 2.5 × 10^-3 M, and the pressure of H2 is 0.8 atm at 298 K.

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To calculate the equilibrium constant  K  for the given reaction, we can use the expression for the reaction quotient  Q  and set it equal to K since the reaction is at equilibrium. The expression for Q is given by:Q = [Zn^2+][H2] /  [H^+]^2 At equilibrium, Q = K. Therefore,K = [Zn^2+][H2] /  [H^+]^2 We are given the concentration of Zn^2+ as 0.25 M, [H^+] as 2.5  10^-3 M, and the pressure of H2 as 0.8 atm. However, we need to convert the pressure of H2 to its concentration using the ideal gas law:PV = nRTWhere P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant  0.0821 L atm/mol K , and T is the temperature in Kelvin. We can rearrange the equation to solve for the concentration  n/V  of H2:Concentration of H2 = P /  RT Plugging in the given values:Concentration of H2 = 0.8 atm /  0.0821 L atm/mol K * 298 K  = 0.0325 MNow we can calculate the equilibrium constant K:K = [0.25][0.0325] /  [2.5  10^-3]^2 K = 0.008125 /  6.25  10^-6 K = 1300The equilibrium constant for the given reaction is approximately 1300.

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