Consider the following reversible reaction:
2 CO(g) + O₂(g) 2CO₂(g). Describe a way to increase the CO concentration at equilibrium. Explain your reasoning.

Answers

Answer 1

The given equilibrium reaction indicates the reaction between carbon monoxide and the oxygen and forms carbon dioxide. In this reaction, by increasing the concentration of the carbon monoxide, the equilibrium shifts towards the right. Since, the reactant concentration increases, the equilibrium stress decreases the concentration of the reactants and therefore, the equilibrium shift towards the right side of the equation.

reaction indicates that carbon monoxide reacts with oxygen and forms carbon dioxide gas.

By increasing the concentration of carbon monoxide image, the equilibrium in the reaction shifts towards the right.

Answer 2

The effect of change of pressure, concentration and temperature on the state of equilibrium can be predicted with the help of a principle known as Le Chatelier's principle. Here by increasing the concentration of CO₂, the amount of CO formation can increase.

What is Le Chatelier's principle?

If a system in equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to reduce or to counteract the effect of change. When the concentration of any substance in equilibrium is increased, the equilibrium shifts in the direction which decreases the concentration of that substance.

On other hand, increasing the concentration of any of the products results in shifting the equilibrium in the backward reaction forming more products. So here increasing the amount of CO₂, shifts equilibrium in the backward direction and forms more amount of CO.

To know more about Le Chatelier's principle, visit;

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Related Questions

A gas cylinder contains 5.0 kg of propane. How many propane molecules are in the cylinder? The Avogadro constant, L = 6.022 x 1023 mol^-1
A 6.8 × 10^22
B 7.2 x 10^22
C 6.8 × 10^25
D 7.2 x 10^25
[1 mark]​

Answers

Answer:

Here's your answer.

M(C3H8) = 44 g/mol

n(C3H8) = 5000 (g)/44 (g/mol) = 113,6 mol

[tex] N = n \times N_{a} = 113,6 \times 6,02 \times {10}^{23} \\ \\ = 684,1 \times 10 ^{23} \sf{molecules} [/tex]

Hope it helps you from my side
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