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Cerrena [4.2K]
2 years ago
9

Explain the impact that changing pressure has on a system in a state of dynamic equilibrium. What will happen when pressure on a

reaction mixture at equilibrium and with fewer moles on the reactant side is increased?
Chemistry
1 answer:
Gelneren [198K]2 years ago
7 0

Answer:

The reverse direction is favored; formation of reactants.

Explanation:

Hello,

In this case, when the chemical reactions are being carried out in gaseous phase, the pressure has a significant effect if the number of moles of reactants differ from those of the products favoring the equilibrium towards the direction having the fewer amount of moles.

In such a way, if the moles of the reactants are less than those of products, increasing the pressure will reverse the reaction towards reactants again as equilibrium shall be reestablished.

Best regards.

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the answer should be D. 462 cm³ 
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COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at
mestny [16]

<u>Answer:</u> The value of K_p for the reaction at 690 K is 0.05

<u>Explanation:</u>

We are given:

Initial pressure of COCl_2 = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

                  COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial:            1                    -         -

At eqllm:       1-x                 x        x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

[(1 - x) + x+ x]=1.2\\\\x=0.2atm

The expression for K_p for above equation follows:

K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}

p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm

Putting values in above equation, we get:

K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05

Hence, the value of K_p for the reaction at 690 K is 0.05

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