Calculate the equilibrium constant for the following reaction at 25 C The correc

Question

Calculate the equilibrium constant for the following reaction at 25 C

The correct answer is e. 3 x 10^57 but I keep getting c. 3 x 10^11.

Calculate the equilibrium constant for the following reaction at 25 degree C, 2 IO^-_3(aq) + 5 Hg(l) + 12 H^+(aq) rightarrow I_2(s) + 5 Hg^2+(aq) + 6 H_2O(l) given the following thermodynamic information. IO^-_3(aq) + 6 H^+(aq) + 5 e^- rightarrow I_2(s) + 3 H_2O(l) E degree = +1.20 V Hg^2+(aq) + 2 e^- rightarrow Hg(l) e degree = +0.86 V 3 times 10^-58 6 times 10^5 3 times 10^11 6 times 10^28 3 times 10^57

Explanation / Answer

The reactions are 2IO3- + 6H+ +5e- ---->I2+ 3H2O, Eo= 1.2V (1)

Reversing the second reaction Hg(l) ------>Hg+2+ 2e-, Eo=-0.86V (2)

Multiplying Eq.1 with 2 and Eq.2 with 5 gives the desired reaction

4IO3- + 12H+ + 5Hg(l) ---->2I2+ 6H2O+ 5Hg+2, EO= 1.2-0.86=0.34V

There are 10 electrons transferred and hence n=10

deltaG=-nFE= -10*96500*0.34 joules=-328100 Joules

since deltaG=-RTlnK=-328100

lnK= 328100/(298*8.314), K= 3.2*1057, ( E is correct)

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