Question 1

An increase in pressure would favour which of the following reaction?

Accepted Answer

In accordance with Le Chatelier's Principle which states that with an increase in pressure, the reaction shifts towards the direction which proceeds with the decrease in the number of moles.In the reaction  2SO_{2}(g) + O_{2}(g)\leftrightharpoons 2SO_{3}(g) , the number of moles decreases in the forward reaction from  3  to  2 . Also, the number of moles increases in the reverse reaction from  2  to  3 .Thus, the correct answer is option  B .

Question 2

The reaction  A + B \leftrightharpoons C + D + heat , has reached equilibrium. It is possible to make the reaction to proceed forward:

Accepted Answer

The reaction  A + B \Leftrightarrow C + D + heat  is an exothermic reaction. According to Le Chatelier's principle, as the temperature is lowered, the reaction will proceed in the direction so as to liberate heat and nullify the effect of lowering the temperature.Hence, the reaction will proceed in the forward direction by lowering the temperature.

Question 3

1 mole of A (g) is heated to  300^{\circ} C  in closed one litre vessel till the following equilibrium is reached:  A(g)\Leftrightarrow B(g) . The equilibrium constant for the reaction at  300^{\circ}C  is 4. What is the concentration of B (in  mol. lit^{-1} ) at equilibrium?

Accepted Answer

Initial concentration of  A  is 1 M. Let  x  represents the concentration of  B  in the equilibrium mixture. The concentration of  A  in the equilibrium mixture will be  1-x . The expression for the equilibrium constant is  K=\dfrac {B} {A} .Substitute values in the above expression. 4=\dfrac {x} {1-x} .Hence,  x=0.8 .

Question 4

A mixture of  0.3  mole of  H_{2}  and  0.3  mole of  I_{2}  is allowed to react in a  10\ litre  evacuated flask at  500^{\circ}C . The reaction is  H_{2} + I_{2} \rightleftharpoons 2HI , the  K  is found to be  64 . The amount of unreacted  I_{2}  at equilibrium is

Accepted Answer

The equilibrium reaction is  H_{2}(g) + I_{2}(g) \Leftrightarrow 2HI(g) .The number of moles of various species present initially and at equilibrium are given in the following table.               H_2              I_2 HI Initial               0.3             0.3   0 change               x               x 2x equilibrium             0.3-x              0.3-x                     2x The expression for the equilibrium constant is  K_c=\dfrac {[HI]^2} {[H_2][I_2]} Substitute values in the above expression. 64=\dfrac {(\dfrac {2x} {10})^2} {(\dfrac {0.3-x} {10}) 	imes (\dfrac {0.3-x} {10})} or  x=0.24 . The amount of unreacted iodine is  0.3-x=0.3-0.24=0.06 .

Question 5

4.5 moles each of hydrogen and iodine are heated in a sealed 10-liter vessel. At equilibrium 3.0 moles of HI were found. The equilibrium constant for   H_{2}(g)  +  I_{2}(g) \rightleftharpoons  2HI(g)    is:

Accepted Answer

The equilibrium reaction is  H_{2}(g) + I_{2}(g) \rightleftharpoons  2HI(g) .The following table lists the equilibrium concentrations of various species.      H_2         I_2        HI Initial     4.54.50Change     -x-x2xEquilibrium     4.5-x4.5-x2xBut for [HI],   2x=3 , So  x=1.5 Hence, the equilibrium number of moles of hydrogen, iodine, and HI are 3,3 and 3 respectively.The total volume is 10 L.Hence, the equilibrium concentrations of hydrogen, iodine, and HI are 0.3M,0.3M and 0.3M respectively.The expression for the equilibrium constant is  K_c= \dfrac {[HI]^2} {[H_2][I_2]} Substitute values in the above expression.  K_c= \dfrac {(0.3)^2} {(0.3)(0.3)}=1 .

Question 6

In the process  N_{2} + 3H_{2} \leftrightharpoons 2NH_{3} , the initial concentration of nitrogen and hydrogen are  1  mole per litre and  3  moles per litre respectively. The equilibrium constant of the reaction is  x . The value of  K_{c}  for  2NH_{3} \leftrightharpoons N_{2} + 3H_{2}  will be:

Accepted Answer

When the first reaction  (N_{2} + 3H_{2} \leftrightharpoons 2NH_{3})  is reversed, second reaction  (2NH_{3} \leftrightharpoons N_{2} + 3H_{2})  is obtained.In such a case, the value of the equilibrium constant for the second reaction will be the reverse of the value of the equilibrium constant for the first reaction.Thus,  K= \dfrac {1} {K_1}=\dfrac {1} {x} .

Question 7

The equilibrium constant for the reaction  A \Leftrightarrow B  is  K . The equilibrium constant for the reaction  mA \Leftrightarrow mB  is:

Accepted Answer

Let  K  is the equilibrium constant for the reaction,  A \Leftrightarrow B .Let  K'  be the equilibrium constant for the reaction,  mA \Leftrightarrow mB . K=\dfrac {B} {A}  or  B=KA  and  K'=\dfrac {B^m} {A^m}. Substituting  B=KA  in the expression for  K' , we get K'=\dfrac {B^m} {A^m}=\dfrac {(KA)^m} {A^m}=K^m 	imes \dfrac {(A)^m} {A^m}=K^m. Hence, option c is the correct answer.

Question 8

A reversible reaction:  SO_{2}(g) + NO_{2}(g) \leftrightharpoons SO_{3}(g) + NO(g)  .. (1) , takes place in two reversible steps as given below with equilibrium constant values  2.0  and  0.45  respectively. SO_{2}(g) + \cfrac{1}{2} O_{2}(g)\leftrightharpoons  SO_{3}(g), K = 2.0......(2) NO_{2}(g)\leftrightharpoons NO(g)+ \cfrac{1}{2}O_{2}(g), K = 0.45......(3) The equilibrium constant  K_{c}  for reaction  (1)  is:

Accepted Answer

The reaction is  SO_{2}(g) + NO_{2}(g) \Leftrightarrow SO_{3}(g) + NO(g) ...  (1) It is obtained by adding the following two reactions: SO_{2}(g) + \frac{1}{2} O_{2}(g)\leftrightharpoons  SO_{3}(g),  K_1 = 2.0 ... (2) NO_{2}(g)\leftrightharpoons  NO(g)+ \frac{1}{2}O_{2}(g),  K _2= 0.45 ... (3) The value of the equilibrium constant of reaction  (1)  is obtained by multiplying the values of the equilibrium constants of the reactions  (2)  and  (3) .Thus,  K=K_1	imes K_2=2.0 	imes 0.45 =0.9 .