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kaheart [24]
3 years ago
10

The reaction h2co3 h2o<-> h3o hco3– takes place in water. what happens to the equilibrium when the pressure is increased?

(1 point)it favors formation of reactants.it favors formation of products.it does not change.it is conserved.
Chemistry
2 answers:
PSYCHO15rus [73]3 years ago
5 0
I think the correct answer from the choices listed above is the first option. When the the pressure is increased, the equilibrium of the reaction would favor <span>formation of reactants. One indication would be the gas that is present in the product side. Increasing the pressure would allow the products to react and form the reactants. Hope this helps.</span>
Anarel [89]3 years ago
5 0

Answer: It does not change.

Explanation:

H_2CO_3(aq) +H_2O\rightarrow H_3O^+(aq)+HCO_3^-(aq)

According to Le Chatelier's principle, if an equilibrium reaction is subjected to a change, the reaction adjusts itself in a way to undo the change imposed.

The effect of pressure affects the equilibrium only when the reactants or products are in gaseous phase.

As none of the reactants or products is in gaseous state, there is no effect of pressure on equilibrium.

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Answer:

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Explanation:

We can divide the distance by 60 seconds to find the trains average rate of speed.

803.98 / 60 ≈ 13.4

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A body that weighs 6 Newton on the earth will weighs – on the moon in newton<br>​
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Which sentence describes the bond formed between two identical atoms?
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3 0
3 years ago
Read 2 more answers
Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe
baherus [9]

Answer:

The equilibrium constant is  K =0.02867

The equilibrium pressure for oxygen gas is  P_{O_2} =  0.09367\  atm

Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

   The Gibbs free energy forM_3 O_4_{(s)} is  \Delta G^o_{1}  = -8.80 \ kJ/mol

  The Gibbs free energy forM{(s)} is  \Delta G^o_{2}  = 0 \ kJ/mol

    The Gibbs free energy forO_2{(s)} is  \Delta G^o_{3}  = 0 \ kJ/mol

The Gibbs free energy of the reaction is mathematically represented as

         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

         \Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)

Substituting values

From the balanced equation

         \Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]

        \Delta G^o_{re} = 8.80 kJ/mol =8800J/mol

The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          K_p =[ P_{O_2}]^{\frac{3}{2} }

Where   [ P_{O_2}] is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

            K =0.02867

Now substituting this into the equation above to obtain the equilibrium of oxygen

           0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}

multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

3 0
3 years ago
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