Chemistry, 11.03.2020 16:44 corbino710

If a reaction mixture initially contains a COCO concentration of 0.1490 MM and a Cl2Cl2 concentration of 0.177 MM at 1000 KK. What is the equilibrium concentration of COCO at 1000 KK?

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Answer from: ahrensy8444

a.) What is the equilibrium concentration of CO at 1000 K? : 0.0130M

b.) What is the equilibrium concentration of Cl₂ at 1000 K?: 0.0410M

c.) What is the equilibrium concentration of COCl₂ at 1000 K?: 0.136M

Explanation:

The complete question is:

Kc= 255 at 1000K for reaction CO(g) + Cl₂(g) ⇄ COCl₂(g). If a mixture initially contains a CO concentration of 0.1490M and a Cl₂ concentration of 0.177M at 1000K, what are the equilibrium concentration of CO, Cl₂ and COCl₂ at 1000K?

a.) What is the equilibrium concentration of CO at 1000 K?

b.) What is the equilibrium concentration of Cl₂ at 1000 K?

c.) What is the equilibrium concentration of COCl₂ at 1000 K?

Solution

1. Write the equilibrium equation

CO(g) + Cl₂(g) ⇄ COCl₂(g)

2. Build the ICE table (initial, change, equilibrium) table

Molar concentrations:

CO(g) + Cl₂(g) ⇄ COCl₂(g)

I 0.1490 0.177 0

C -x -x +x

E 0.01490 - x 0.177 - x x

3. Write the equation of the constant of equilibrium

$K_c=\dfrac{[COCl_2]}{[CO][Cl_2}}$

$255=\dfrac{x}{(0.1490-x)(0.177-x)}$

4. Solve the equation

255(0.026373-0.1490x-0.177x+x^2)=x

6.725115-37.995x-45.135x+255x^2=x

255x^2-84.13x+6.725115=0

Use the quadratic equation to obtain:

x = 0.136 and x = 0.194

It cannot be greater than 0.1490, thus the correct solution is 0.136

And the concentrations are:

a.) Concentration of CO: 0.1490M - 0.136M = 0.0130M

b.) Concentration of Cl₂: 0.177M - 0.136M = 0.0410M

c.) Concentration of COCl₂: 0.136M

Answer from: Quest

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh $pcl_{3}$ +δh $cl_{2}$ ] - [δh $pcl_{5}$ ]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

Answer from: Quest

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