Question
questions about equilibrium constant and how to make energy calculations. just 12 and 13 please. I feel like this is far more simple than i am making it out to be
Il You are given pure samples of butane CH,CH2CH CH and 2-butene CH,CH CHCH what prediction would you make concerning their standard molar entropies at 298 K? D) More information is needed to make reasonable predictions 2. Consider the reaction: Calculate K, at 25 C given the following thermodynamic data Substance AG (kJ/mol) -6.075 Cl (g) 19.38 A 3.50 x 10 B) 3.36 x 3500 D) 3360 13. The equilibrium constant for the reaction: 2Fe (aq) Hea (aq) 2Fe" (aq) +2Hg "(aq) is K -9l x 10' at 298 K. What is AG at this temperature? A) 287 x B) 29 x 10 Jmol C) 30000 J/mol D) 39 x 10 Jmol 4. Consider the following balanced redox reaction 16H (aq) +2Mno (aq) 10CI(aq) 2Mn (aq) +5Cl (z) +8H1O(I). Which species is being redu B) H+ D) Mimo. 5.A voltaic cell has the following cell notation: Zn(s) l zniraq) (aq)li graphite Which of the following represents the correctly balanced spontaneous reaction equation for the D) (5) +Zn (aq) 21(aq) Zn(s) 6, What is the coefficient of NO, (aq) in the following redox reaction when it is correctly balanced? Sbos) +NO,(aq) Sb OMs) +NOKg) acidic
Explanation / Answer
Equilibrium constant
(12)
dGof(rxn) = dGof(products) - dGof(reactants)
= (2 x -6.075) - (19.38)
= -31.53 kJ/mol
Now,
dGo = -RTlnKp
31530 = 8.314 x 298 lnKp
Kp = 3.36 x 10^5
Answer : B) 3.36 x 10^5
(13) with Kc = 9.1 x 10^=6
dG = -RTlnKp
= -8.314 x 298 ln(9.31 x 10^-6)
= 2.9 x 10^4 J/mol
Answer : B) 2.9 x 10^4 J/mol
