When does chemical reaction proceed spontaneously? The Second Law gives the answer. Consider a syste — Physical Chemistry — Thermodynamics Chemistry Question
The Second Law of thermodynamics applied to a chemical reaction
When does chemical reaction proceed spontaneously? The Second Law gives the answer.
Consider a system, a chemical reactor in Fig. 1. Pressure р and temperature Т inside the reactor are kept constant. There is no transfer of matter into or out of the system.
А + В ⇄ С
p = const
Т, nA,nB = const
Fig. 1. Chemical reaction inside a reactor
According to the Second Law, every spontaneous process in such a reactor leads to the decrease of the Gibbs free energy, Gsystem, i.e. ∆Gsystem < 0.
If the chemical reaction, e.g. А + В = С (a), is the only process inside the reactor
∆Gsystem = ∆Greaction(a) ∆ξ(a)
∆ξ(a) = ∆nC = -∆nA = -∆nB (1)
where ∆Greaction and ∆ξ are the Gibbs free energy and the extent of reaction (a), respectively, ∆nA, ∆nB, ∆nC are changes of the numbers of moles of A, B, C in the reaction (a).
Relate ∆ξ to ∆ni of reactants and products of the following reaction
1/6 C6H12O6 + O2 = CO2 + H2O (a)
Model Answer
∆nH2O = ∆nCO2 = -∆nO2 = – 6 ∆nC6H12O6 = ∆ξ
Prove that, according to the Second Law, ∆Greaction < 0 for any single spontaneous chemical reaction in the reactor (Fig.1).
The Gibbs free energy of the chemical reaction (a) is:
∆Greaction = ∆G0reaction + RT ln([C] / ([A][B])) < 0 (2)
where [C], [A], [B] are time variant concentrations inside the reactor in the course of spontaneous reaction. Using the law of mass action, relate ∆Greaction to the ratio of rates of forward r1 and reverse r–1 reaction (a). Consider both reactions as elementary ones.
Model Answer
If spontaneous chemical reaction is the only process in the reactor, ∆Gsystem < 0. The value of ∆ξ for spontaneous reaction is positive, ∆ni are positive for the products and are negative for the reactants (minus sign makes them positive!). Thus,
∆Gsystem = ∆Greaction ∆ξ < 0.
Since both forward and reversed reactions are elementary ones, the following equality may be written at equilibrium:
r1eq = k1[A]eq[B]eq = r–1eq = k–1[C]eq
and
K = k1 / k–1 = [C]eq / ([A]eq[B]eq)
[C]eq, [B]eq, [A]eq are concentrations at equilibrium, K is the equilibrium constant of the reaction. Making use of the well-known formula ∆G0 = – RT ln K
from equation (2) one gets
∆G = –RT ln K + RT ln([C]/[A][B]) = RT ln(k–1/k1) + RT ln(r–1k1 / r1k–1) = RT ln(r–1/r1) (3)
Derive the expression (2) for ∆Greaction of the following chemical transformations:
(a') H2(g) + Br2(g) = 2 HBr(g)
(a'') H(g) + Br2(g) = Br(g) + HBr(g)
(a''') CaCO3(s) = CaO(s) + CO2(g)
Model Answer
∆Greaction(a') = ∆G0reaction + RT ln([HBr]2 / ([H2][Br2]))
∆Greaction(a'') = ∆G0reaction + RT ln([HBr][Br] / ([H][Br2]))
∆Greaction(a''') = ∆G0reaction + RT ln[СO2]
For which of these reactions the relation between ∆Greaction and r1, r–1 derived in Problem 2 is valid?
Model Answer
Equation (3) may be used only in case (a''). Other two reactions are not elementary ones.
The observed rate of chemical reaction, robs, is defined as robs = r1 – r–1.
Let reaction (a) proceed spontaneously. At a certain moment
robs / r1 = 0.5, [A] = 0.5 mol dm-3, [B] = 1 mol dm-3, [C] = 2 mol dm-3.
Find the equilibrium constant, K, of the reaction (a), T=298K.
Model Answer
robs / r1 = (r1 – r–1)/r1 = 1 – r–1/r1 = 0.5; r–1/r1 = 0.5
r–1/r1 = (k–1[C]) / (k1[A][B]) = 1/K * 2/(0.5×1) = 0.5; K = 8 mol–1 dm3
Plot robs as a function of
a) r1 at ∆GReaction = const;
b) r1 at r–1 = const;
c) ∆GReaction, at r1 = const.
Which thermodynamic and kinetic parameters of a chemical reaction are influenced by a catalyst? Put plus (+) into the cell of the Table if a catalyst may cause a change of the corresponding parameter, (–) otherwise.
Table
robs r1 r1/r–1 ∆Greaction robs/r1
Model Answer
Table
robs: +
r1: +
r1/r–1: –
∆Greaction: –
robs/r1: –