Question

I cannot seem to figure it out.

The standard Gibbs-free energy of a system is related to its equilibrium constant through the following equation. AG = R.T.In(K) In this equation R is the gas constant, T is the temperature, and the next to AG defines the conditions as standard ambient temperature and pressure, i.e. "SATP". (Answer the following questions to three significant figures.) (a) Given an equilibrium constant of 6.28 x 10-3, what is its standard Gibbs-free energy? 4.9 12.6 kJ/mol (b) Given a standard Gibbs-free energy of 34.4 kJ/mol, what is the value of its equilibrium constant? 4.9 9.40e-7 X (c) Suppose a reaction has a standard Gibbs-free energy of -0.762 kJ/mol, calculate the relative ratio of product to reactant. product reactant 4.9) 42.37 x % 49 57.63 x % Submit Answer Practice Another Version

Answer 1

Keq= 24 = fr&xto 34.4 kamol odno 34.4 xlooo 7 moly Geren equation is • Doro PT enkeq. lord pe where Ra Gas constant = 8.314 II molt TTemperature = 273K A SAP 30-U YI000 Imol - 8.319 JKMDTX973 One I 634-4 X10oodmolt' - 8.314 Amor-x27301 enk - 3440 2869072 enk = -15.1560 Takis anti - logarithm nu. anti (-15.1560) c olo kegi = 2.62x107 | Answer - DG = using 0-762 kolmo = 762 Aq = - RT en keq dus Imoly mano

keq = antilog E7 762 Jmolt --8-314 X 273k IMOINE keq - antio 4.762-12] keq = antilog Co-33573 [keq = 1.398] since keq = [products} [Read ants}. 1.398 = [product TReactant] - Product = 2X1348 React ant]. Total amount = (Product] + [Reactants. = 2 [Reactan) + treacan) - a reactants (9+1) = 3 Creactant af Product BCReactant) xido Product = X100 = 8 Total 3 Reactant] - 66.67% 7 of Reactant Creact aut] xloo = [Reactants total 3x (Reactant - 4X100 = 33-334. Products Reactant = 66-67 : 33-3.2 Answer