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Page No 227: - Chapter 7 Equilibrium class 11 ncert solutions Chemistry - SaraNextGen [2024]


Question 7.25:

Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(a) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8125/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_1d21feb5.gif

(b) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8125/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m29cf4b37.gif

(c) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8125/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m44cd17d3.gif

Answer:

(a) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(b) The number of moles of reaction products will decrease.

(c) The number of moles of reaction products remains the same.

Question 7.26:

Which of the following reactions will get affected by increasing the pressure?

Also, mention whether change will cause the reaction to go into forward or backward direction.

(i) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8126/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m6a01ce0f.gif

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8126/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_2218ef3.gif

(iii) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8126/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m3ae4924c.gif

(iv) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8126/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m3ac0f0fd.gif

(v) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8126/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_42655eb5.gif

(vi) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8126/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m3e1609a6.gif

Answer:

The reactions given in (i), (iii), (iv), (v), and (vi) will get affected by increasing the pressure.

The reaction given in (iv) will proceed in the forward direction because the number of moles of gaseous reactants is more than that of gaseous products.

The reactions given in (i), (iii), (v), and (vi) will shift in the backward direction because the number of moles of gaseous reactants is less than that of gaseous products.

Question 7.27:

The equilibrium constant for the following reaction is 1.6 ×105 at 1024 K.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8127/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_19e65bc.gif

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Answer:

Given,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8127/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_b6c25aa.gif for the reaction i.e., https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8127/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_4c6189ed.gif

Therefore, for the reactionhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8127/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m5f2b95a2.gif  the equilibrium constant will be,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8127/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m81cb48c.gif

Now, let p be the pressure of both H2 and Br2 at equilibrium.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8127/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_6974de9.gif

Now, we can write,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8127/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_375dfda0.gif

Therefore, at equilibrium,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8127/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m3b12a18a.gif

Question 7.28:

Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8128/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m754a25dd.gif

(a) Write as expression for Kp for the above reaction.

(b) How will the values of Kp and composition of equilibrium mixture be affected by

(i) Increasing the pressure

(ii) Increasing the temperature

(iii) Using a catalyst?

Answer:

(a) For the given reaction,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8128/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m4f4eecec.gif

(b) (i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

Question 7.29:

Describe the effect of:

a) Addition of H2

b) Addition of CH3OH

c) Removal of CO

d) Removal of CH3OH

on the equilibrium of the reaction:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8129/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_306a7c57.gif

Answer:

(a) According to Le Chatelier’s principle, on addition of H2, the equilibrium of the given reaction will shift in the forward direction.

(b) On addition of CH3OH, the equilibrium will shift in the backward direction.

(c) On removing CO, the equilibrium will shift in the backward direction.

(d) On removing CH3OH, the equilibrium will shift in the forward direction.

Question 7.30:

At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 ×10-3. If decomposition is depicted as,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8130/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_m6441e4dc.gif  ΔrH° = 124.0 kJmol–1

a) Write an expression for Kfor the reaction.

b) What is the value of Kc for the reverse reaction at the same temperature?

c) What would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased? (iii) The temperature is increased?

Answer:

(a) https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8130/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_23f00f81.gif

(b) Value of Kc for the reverse reaction at the same temperature is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/200/8130/NS_06-11-08_Utpal_11_Chemstry_7_73_GSX_SG_html_5fb03479.gif

(c) (i) Kwould remain the same because in this case, the temperature remains the same.

(ii) Kis constant at constant temperature. Thus, in this case, Kwould not change.

(iii) In an endothermic reaction, the value of Kincreases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kwill increase if the temperature is increased.

Also Read : Page-No-228:-Chapter-7-Equilibrium-class-11-ncert-solutions-Chemistry

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