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Calculate the standard enthalpy change for the reaction A(s) + 2B(g) → 3C(l) + D(g) given the following data:• ΔHf° for A(s) is -100 kJ/mol• ΔHf° for C(l) is -50 kJ/mol• ΔS° for the reaction is 450 J/K•mol• ΔS° for D(g) is 200 J/K•mol

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To calculate the standard enthalpy change  H  for the reaction, we need to use the following equation:H =  Hf products  -  Hf reactants First, we need to find the Hf for D g . We can use the given S values and the equation:G = H - TSSince the reaction is at standard conditions, the temperature  T  is 298 K. We know that at equilibrium, G = 0. Therefore, we can rearrange the equation to find H for D g :H = TSH for D g  =  298 K  200 J/Kmol  = 59600 J/mol = 59.6 kJ/molNow we can calculate the standard enthalpy change for the reaction:H = [3Hf C  + Hf D ] - [Hf A  + 2Hf B ]Since we don't have the Hf for B g , we can't directly calculate the enthalpy change for the reaction. However, we can express the enthalpy change in terms of Hf B :H = [3 -50 kJ/mol  + 59.6 kJ/mol] - [-100 kJ/mol + 2Hf B ]H = [-150 kJ/mol + 59.6 kJ/mol] - [-100 kJ/mol + 2Hf B ]H = -90.4 kJ/mol + 100 kJ/mol - 2Hf B H = 9.6 kJ/mol - 2Hf B Without the Hf for B g , we cannot provide a numerical value for the standard enthalpy change of the reaction. The answer is:H = 9.6 kJ/mol - 2Hf B

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