Question

Consider the reaction:

Zn(s) + 2 H+(aq) = Zn2+(aq) + H2(g)

At 25C, calculate:

a) ∆G˚ for the reaction, given that: ∆Gf Zn(s) = 0, ∆Gf(H+) = 0, ∆Gf(H2) = 0, ∆Gf(Zn2+) = -147.1 kj/mol

b) ∆G, when P(h2) = 750 mmHg, [Zn2+ aq] = 0.10 M, [H+] = 1.0 x 10^-4 M

c) The pH when ∆G - -100 kJ, P(h2) = 0.922 atm, [Zn2+] = 0.200 M and the mass of Zn is 155 g.

Answer 1

Zn(s) + 2 H+(aq) = Zn2+(aq) + H2(g)

∆G˚ for the reaction, given that: ∆Gf Zn(s) = 0, ∆Gf(H+) = 0, ∆Gf(H2) = 0, ∆Gf(Zn2+) = -147.1 kj/mol

deltaG0= ∆Gf(Zn2+)*1 + deltaGfH2*1- { deltaG(Zn*1+ deltaG(H+)*2}

1, 1,1,2 are coefficients of products and reactants in the reaction

deltaGo= -147.1 KJ/mol

2. Ph2= 750 mm Hg= 750/760 atm =0.99 atm, [Zn2+ aq] = 0.10 M, [H+] = 1.0 x 10^-4 M

Zn(s) + 2 H+(aq) = Zn2+(aq) + H2(g) , Q = [Zn+2] [PH2]/ [[H+]² =0.1*0.99/(1*10^-4)²= 9.9*10⁶

deltaG= deltaG0+RTln Q ,= -147.1*1000+8.314*298*ln(9.9*10⁶)= -129768 Kj/mole

3. Moles of Zinc in 155 gm = mass/ atomic weight =155/65.38 =2.4

deltaH= -100/2.4 Kj/mole=-41.7 Kj/mole= -41.7*1000 = delltaGO+RT*lnQ

-41.7*1000 = -147.1*1000+8.314*298*lnQ

105400= 8.314*298*lnQ

lnQ= 105400/(8.314*298)= 42.54

Q= 2.9*10¹⁸

Q = [Zn+2] [PH2]/ [[H+]²

2.9*10¹⁸⁼ 0.2* 0.922/[H+]²

[H+]2 = 0.2*0.922/(2.9*10¹⁸). [H+]= 2.52*10^-10, pH= 9.6

Q=