The equilibrium constant, K₂, for the following reaction is 0.636 at 600 K.Calculate Ke for this reaction at this temperature.COC1₂ (g) → CO(g) + Cl₂ (g)Kc=

Answered: The equilibrium constant, Kp, for the…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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The equilibrium constant, K₂, for the following reaction is 0.636 at 600 K.
Calculate Ke for this reaction at this temperature.
COC1₂ (g) → CO(g) + Cl₂ (g)
Kc
=

Expert Solution

Step 1

k_p is the equilibrium constant where the concentrations of gaseous reactants and products are expressed by pressure P.

k_C is the equilibrium constant where the concentrations of gaseous reactants and products are expressed by molarity M.

According to ideal gas equation, PV = nRT (P- pressure of the components)

$P = (\frac{n}{V}) R T$ (V- volume of components)

$P = C R T$ (C- concentration) (n - no of moles of components)

The reaction is (R- universal gas constant = 0.082 L.atm mole^-1K^-1)

COCl₂(g) $\to$ CO(g) + Cl₂(g)

According to question k_p = 0.636 at T = 600K

$k_{p} = \frac{{P^{1}}_{C O} \times {P^{1}}_{C l_{2}}}{{P^{1}}_{C O C l_{2}}} ∵ c o e f f i c i e n t o f C O C l_{2}, C O, a n d C l_{2} a r e 1 = \frac{{(C_{C O} x R T)}^{1} \times {(C_{C l_{2}} x R T)}^{1}}{{(C_{C O C l_{2}} \times R T)}^{1}} = (\frac{C_{C O} \times C_{C l_{2}}}{C_{C O C l_{2}}}) \times {(R T)}^{1} = k_{C} x {(R T)}^{1} k_{C} = \frac{k_{p}}{R T} = \frac{0.636 a t m}{0.082 L . a t m m o l e^{- 1} K^{- 1} x 600 K} = 0.0129 m o l e / L$

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