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Calculate the standard enthalpy change (ΔH°), standard entropy change (ΔS°) and standard Gibbs free energy change (ΔG°) of the reaction:2A(g) + 3B(g)  → C(g) + 2D(g)given thatΔH°f[A(g)] = 200 kJ mol^-1Δ H°f [B(g)] = 100 kJ mol^-1Δ H°f [C(g)] = 400 kJ mol^-1Δ H°f [D(g)] = 0 kJ mol^-1ΔH°= -400 kJ mol^-1Temperature (T) = 298 KThe student needs to use their knowledge in quantum chemistry and thermodyna

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To calculate the standard enthalpy change  H  for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants For this reaction, we have:H = [ 1 mol C  400 kJ/mol  +  2 mol D  0 kJ/mol ] - [ 2 mol A  200 kJ/mol  +  3 mol B  100 kJ/mol ]H = [400 -  400 + 300 ] = -300 kJ/molNow, we need to find the standard entropy change  S  for the reaction. Since we are not given the entropy values for each species, we will assume that the given H value is correct and use the following equation to find S:G = H - TSWe are given H = -400 kJ/mol and T = 298 K. We need to convert H to J/mol:H = -400 kJ/mol * 1000 J/kJ = -400,000 J/molNow, we can rearrange the equation to solve for S:S =  H - G  / TS =  -400,000 J/mol -  -300,000 J/mol   / 298 KS =  100,000 J/mol  / 298 KS = 335.57 J/mol KFinally, we can calculate the standard Gibbs free energy change  G  for the reaction using the given H value and the calculated S value:G = H - TSG = -400,000 J/mol -  298 K  335.57 J/mol K G = -400,000 J/mol - 100,000 J/molG = -500,000 J/molSo, the standard enthalpy change  H  is -300 kJ/mol, the standard entropy change  S  is 335.57 J/mol K, and the standard Gibbs free energy change  G  is -500 kJ/mol.

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