Chemical Equilibrium: Concept, Extent of Reaction, and Equilibrium Constant - Prof. Stefan, Study notes of Physical Chemistry

Download Chemical Equilibrium: Concept, Extent of Reaction, and Equilibrium Constant - Prof. Stefan and more Study notes Physical Chemistry in PDF only on Docsity! Chemical Equilibrium One of the most fundamental applications of thermodynamics is to chemical reactions at equilibrium. The underlying fundamental idea that we describe here is that ΔG = 0 for a system in equilibrium (at constant temperature and pressure), and the sign of ΔG determines whether or not a given process or chemical reaction will occur spontaneously at constant T and P. Consider a general gas phase chemical reaction νAA (g) + νBB (g) ⇔ νyY (g) + νZZ (g) We define a quantity ξ, called the extent of reaction such that the numbers of moles of reactants and products are given by nA = nA0 - νAξ nB = nB0 - νBξ nY = nY0 - νYξ nZ = nZ0 - νZξ where nj0 is the initial number of moles for each species. According to the equations above ξ must have units of moles. As the reaction proceeds from reactants to products, ξ varies from 0 to some maximum value dictated by the stoichiometry of the reaction. Differentiation of the equations gives that rate of change of the number of moles. Reactants dnA = -νAdξ dnB = -νBdξ Products dnY = νYdξ dnZ = νZdξ The negative signs indicate that the reactants are dissappearing and the positive signs indicate that the products are being formed as the reaction progresses from reactants to products. Now consider a system containing reactions and products dG = -SdT + VdP + μAdnA + μBdnB + μYdnY + μZdnZ At constant T and P we have dG = μAdnA + μBdnB + μYdnY + μZdnZ or using the above expression dG = -μAνAdξ - μBνBdξ + μYνYdξ + μZνZdξ or dG = (μYνY + μZνZ -μAνA - μBνB)dξ or (∂G/∂ξ) = (μYνY + μZνZ -μAνA - μBνB) We define the standard free energy of a reaction as (∂G/∂ξ). In other words, (∂G/∂ξ) = ΔrG. As we have seen the units of ΔrG are J/mole. The quantity ΔrG has meaning only if the balanced chemical equation is specified. If all of the gases in the reaction are ideal then μj(T,P) = μjo(T) + RT(lnPj/Po) and substituting this expression into the above equation we find or ΔrG = ΔrGo + RT lnQ where ΔrGo = μYoνY + μZoνZ -μAoνA - μBoνB and The quantity ΔrGo is the standard Gibbs energy for the reaction between unmixed reactants in their standard states at temperature T and a pressure of one bar to form unmixed products in their standard states at the same T and P. The quantity Q is called the reaction quotient. Its magnitude is dependent on the quantity of reactant and product at any given point during the chemical reaction. The pressures in Q are all referenced to Po = 1 bar which ΔrG = νYμYo(T) + νZμZo(T) – ν Aμ Ao (T) – ν BμBo(T) + RT νYln PY PY o + νYln PZ PZ o – νYln PA PA o – νYln PB PB o Q = PY PY o νY PZ PZ o νZ PA PA o ν A PB PB o ν B Although the pressure appears in this expression, the equilibrium constant K(T) is not a function of pressure. Rather it is ξ that is a function of pressure. The greater the total pressure, the smaller ξ must be to compensate. This is an example of Le Chatelier's principle: If a chemical reaction at equilibrium is subjected to a change in conditions that displaces it from equilibrium, then the reaction adjusts toward a new equilibrium state in such a way as to minimize the effect of the change. In other words, if we increase the total pressure, the equilibrium shifts in the direction of fewer moles of total components (toward the left in the present reaction). This minimizes the total pressure, thus acting to minimize the effect of the change. Example: we consider once again the reaction NH3 (g) = 3/2 H2 (g) + 1/2 N2 (g) for which K(T) = 1.36 x 10-3 at 298 K. Determine the extent of reaction x at equilibrium at a pressure of 1millibar and at a pressure of 1 bar. SOLUTION: We construct a table NH3 (g) 3/2 H2 (g) 1/2 N2 (g) 1 - ξ 3/2ξ 1/2ξ The total number of moles is 1 - ξ + 3/2ξ + 1/2ξ = 1 + ξ. Dalton's law tells that the partial pressure of each gas is NH3 (g) 3/2 H2 (g) 1/2 N2 (g) (1 - ξ)P/(1+ξ) 3/2ξP/(1+ξ) 1/2ξP/(1+ξ) K(T) = ξ 2P 1 – ξ2 We substitute these expressions into the equilbrium constant. Now we solve for ξ As P gets large we can see that ξ gets small. This is in accord with Le Chatelier's principle since the number of moles of all components will be reduced as the equilibrium is shifted to the left. If P = 1 mbar then And if P = 1 bar Given the relation ΔrGo = -RT ln K(T) it is also possible to calculate equilibrium constants from tabulated standard molar Gibbs energy data. We simply invert the relation to find K(T) = exp{ -ΔrGo/RT }. We can plot the Gibbs energy, G versus the extent of reaction, ξ. This plot will pass through a minimum at the equilibrium K(T) = PH2 3/2 PN 2 1/2 PNH3 = 3/2ξP 1 + ξ 3/2 1/2ξP 1 + ξ 1/2 (1 – ξ)P 1 + ξ = 3 2 3/2 1 2 1/2 ξP 1 + ξ 2 (1 – ξ)P 1 + ξ = 3 2 3/2 1 2 1/2 ξ2P (1 – ξ2) ξ = KK + 0.569P ξ = 1.36 × 10 – 3 1.36 × 10– 3 + 0.569 0.001 = 0.839 ξ = 1.36 × 10 – 3 1.36 × 10– 3 + 0.569 1 = 0.049 composition or extent of reaction. This is treated concretely by considering the thermal decomposition of N2O4 (g) N2O4 (g) = 2 NO2 (g) Starting initially with an idealized one mole of N2O4, as the reaction proceeds the number of moles will be 1 - ξ of N2O4 and 2ξ of NO2. The Gibbs energy of the reaction mixture is given by G(ξ) = (1 - ξ)GN2O4 + 2ξGNO2 = (1 - ξ)GoN2O4 + 2ξGoNO2 + (1 - ξ)ln PN2O4 + 2ξln PNO2 The reaction is carried out at a constant total pressure of one bar (i.e. PTotal = 1). Therefore, P N2O4 = x N2O4 P Total = x N2O4 P NO2 = x NO2 P Total = x NO2 and we can calculate the mole fractions as we did above by determining the total number of moles present for any given extent of the reaction as (1 - ξ) + 2ξ = 1 + ξ. Choosing the standard states such that ΔfGo N2O4 = Go N2O4 and ΔfGo NO2 = Go NO2, the free energy expression becomes Substituting in the values ΔfGo N2O4 = 97.787 kJ/mole ΔfGo NO2 = 51.258 kJ/mole RT = 2.479 kJ/mole at 298 K the equation becomes In principle, we could find the minimum of this function by taking the derivative with respect to ξ and setting the result equal to zero. This is very messy. An easy way to see that this function passes through a minimum is to plot it. A plot of the function is shown below. xN2O4 = 1 – ξ 1 + ξ , xNO2 = 2ξ 1 + ξ G ξ = 1 – ξ ΔrGN2O4 o + 2ξΔrGNO2 o + 1 – ξ RT ln 1 – ξ1 + ξ + 2ξRT ln 2ξ 1 + ξ G ξ = 1 – ξ 97.787 + 2ξ51.258 + 2.479 1 – ξ ln 1 – ξ1 + ξ + 2ξ ln 2ξ 1 + ξ