chem 102 help please 6 24, what is K ifGo-18,0 kJ for a reaction at 25 °C? A) 7.

Question

chem 102 help please

6 24, what is K ifGo-18,0 kJ for a reaction at 25 °C? A) 7.3 x 10.4 B) 1.2 x 10 C) 1.4 x 10 D) 8.1 x 10 25. Based on the following information Ch(g) +2e--+2 Cl-(aq) Mg2"(aq) + 2 e- 2 Mg(s) E 1.36V E"--2.37 V which of the following chemical species is the strongest oxidizing agent? A) Mg (s) B) CI (aq) A voltaic cell is set up using the system shown below: C) Cl2(g) D) Mg (aq) 26. Ni Ni Cu Cu Which reaction would occur at the cathode? A) B) Cu--, C112+ + 2 e Cu2+ +2 e-Cu D) Ni2+ + 2 e- Ni 27. For the following reaction at 25 °C 12 (g) +Cl: (g) 2ICI(g) H.-26.9 kJ and A"-1 1.3 J K. Calculate Go for this reaction at 25 oC. A) 3390 kJ B) -30.3 k.J C) 102 k D) 50.6 kJ

Explanation / Answer

24) use Go = -2.303 RT log[K]

{ R = gas constant (8.314 J/mol K), T = temperature (298 K), and K is equillibrium cont. }

-18000 J = -2.303 x 8.314 x 298 log [K]

log[K] = 3.154

K = 1425.6 ~ 1.4x103

option c is correct

25) metal which have low standard reducing potentials value are strong oxidising agent

SRP value of Mg2+ < Cl2

so Mg2+ is strong oxidising agent

26) reaction will be

Ni + Cu2+ -----> Ni2+ + Cu

cathode = Cu2+ + 2e- ---> Cu

anode = Ni ---> Ni2+ + 2e-

27) we use gibbs energy equation

Go = H° - TS°

so, simply put value and find G°

G° = -26900 J - [ 298 K x 11.3 J/K ]

G° = -30267.4 J ~ -30.3 KJ

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