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jasenka [17]
3 years ago
12

Indicate whether either an increase or decrease of pressure obtained by changing the volume would increase the amount the produc

t in each of the following reactions a. CO(g) + 2 H_2(g) doubleheadarrow CH_3OH(g) b. 2SO_2(g) +O_2(g) doubleheadarrow 2SO_3(g) c. 2HO_2(g) doubleheadarrow N_2O_4(g)
Chemistry
1 answer:
vazorg [7]3 years ago
5 0

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle:

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On an increase in the volume :

If the volume of the container is increased, the pressure will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where increase in pressure is taking place.And equilibrium will shift in direction where number of gaseous moles are more.

On an decrease in the volume :

If the volume of the container is decreased , the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decease in pressure is taking place.And equilibrium will shift in direction where number of gaseous moles are less.

a. CO(g) + 2 H_2(g)\rightleftharpoons CH_3OH(g)

On an increase in the volume :

As the number of moles of gas molecules is greater at the reactant side. So, the equilibrium will shift in the left direction. So in order to  increase the amount of product we have to increase the pressure of the reactants molecules decrease the volume.

b.2SO_2(g) +O_2(g) \rightleftharpoons 2SO_3(g)

As the number of moles of gas molecules is greater at the reactant side. So, the equilibrium will shift in the left direction. So in order to  increase the amount of product we have to increase the pressure of the reactants molecules decrease the volume.

c. 2NO_2(g)\rightleftharpoons N_2O_4(g)

As the number of moles of gas molecules is greater at the reactant side. So, the equilibrium will shift in the left direction. So in order to  increase the amount of product we have to increase the pressure of the reactants molecules and decrease the volume .

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Expression for the given decomposition reaction is as follows.

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Let us assume that x concentration of N_{2}O_{4} is present at the initial stage. Therefore, according to the ICE table,

                    N_{2}O_{4} \rightarrow 2NO_{2}

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Equilibrium : (x - 0.1)             0.2

Now, expression for K_{p} of this reaction is as follows.

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Putting the given values into the above formula as follows.

          K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

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This means that P_{N_{2}O_{4}} = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

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