part a CO(g)+Cl2(g)=COCl2(g) Carbon monoxide and chlorine gas are allowed to rea

Question

part a
CO(g)+Cl2(g)=COCl2(g)

Carbon monoxide and chlorine gas are allowed to react in a sealed vessel at 469C . At equilibrium, the concentrations were measured and the following results obtained:
CO .710
Cl2 1.19
COCl2 .210


what is the equlilibrum constant Kp of this reaction


part b
Reaction at 775 C
H2(g) +I2(g)=2HI(g)
Initially, only H2 and I2 were present at concentrations of [H_2]= 3.10M and [I_2]= 2.10M. The equilibrium concentration of I_2 is 0.0400 M . What is the equilibrium constant, K_c, for the reaction at this temperature?

Explanation / Answer

Cell Potential of the reaction and equilibrium constant are related as: E° = R·T/(n·F)·ln(K) K = e^(n·F·E°/ (R·T) ) (R universal gas constant, T absolute temperature, n number of electrons exchanged in reaction, F Faraday constant) At T = 298K F/(R·T) = 96485C/mol/ (8.3145J/molK · 298K) = 38.941V?¹ => K = e^(38.941V?¹ · n · E° ) a. Cu²?(aq) + 2e? ? Cu(s) ; E° = 0.340V Ag?(aq) + e? ? Ag(s) ; E° = 0.7996V => Cu(s) ? Cu²?(aq) + 2 e? ; E° = -0.34V 2Ag?(aq) + 2e? ? Ag(s) ; E° = 2·7996V = 1.5992V => Cu(s) + 2 Ag?(aq) ? Cu²?(aq) + 2 Ag(s) ; E° = -0.340V - 1.5992.4V = 1.2592V => K = e^(38.941V?¹ · 2 · 1.2592V ) = e^( 98.069) = 3.90×104² b. Ce4?(aq) + e? ? Ce³?(aq) ; E° = 1.44V Bi³?(aq) + 3e? ? Bi(s) ; E° = 0.32V => 3Ce4?(aq) + 3e? ? 3Ce³?(aq) ; E° = 3·1.44V = 4.32V Bi(s) ? Bi³?(aq) + 3e? ; E° = -0.32V => 3Ce4?(aq) + Bi(s) ? 3Ce³?(aq) + Bi³?(aq) E° = 4.32V - 0.32.4V = 4.0V => K = e^(38.941V?¹ · 3 · 4.0V ) = e^( 467.3 )

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