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OlgaM077 [116]
3 years ago
12

Onsider the following reaction at equilibrium:

Chemistry
1 answer:
11111nata11111 [884]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.

For the given equation:

C(s)+H_2O(l)\leftrightharpoons CO(g)+H_2(g)

A. C is added to the reaction mixture.

If the concentration of reactant specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of reactant specie occurs. So, on increasing C ,equilibrium will shift in right direction.

B. H_2O is condensed and removed from the reaction mixture.

If the concentration of reactant specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of reactant specie occurs. So, on removing water vapor ,the equilibrium will shift in left direction.

C. CO is added to the reaction mixture.

If the concentration of product specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of product specie occurs. So, on increasing CO, the equilibrium will shift in left direction.

D. H_2 is removed from the reaction mixture.

If the concentration of product specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of product specie occurs. So, on removing hydrogen , the equilibrium will shift in right direction.

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What is the pH of pure water at 40.0°C if the Kw at this temperature is 2.92 - 10-147 OA) 6.767 O B) 0 465 O c) 7.000 OD) 7 233
Sergeeva-Olga [200]

Answer:

PH= 6.767     (answer is the A option)

Explanation:

first we need to correct the value in Kw at this temperature is 2.92*10^-14

so, in this case we have that:

Kw=2.92*10^-14 M²

[ H3O^+] [ H3O^+]

[H_{3}O^{+}  ] [OH^{-}  ] = Kw = 2.92*10^{-14} M^{2}   \\\\

at 40ºC

[H_{3}O^{+}  ] = [OH^{-}  ]

[H_{3}O^{+}  ]^{2} = 2.92*10^{-14} M^{2}

[H_{3}O^{+}  ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M

PH= -log10[H_{3}O^{+}  ] = -log10(1.71*10^{-7} ) = 6.767

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