IISER Chemistry - Chemical Equilibrium
Exam Duration: 45 Mins Total Questions : 30
Which one of the following is a reversible process ?
- (a)
mixing of gases by diffusion
- (b)
dissociation of a solute in pure solvent
- (c)
expansion of a gas in vacuum
- (d)
melting of ice without rise in temprature.
(a), (b) and (c) are irreversibles and (d) is reversible as ice and water both can exist at 00C.
In which of the following does the reaction go almost to completion?
- (a)
\(K_{c}=10^{3}\)
- (b)
\(K_{c}=10^{2}\)
- (c)
\(K_{c}=10^{-2}\)
- (d)
\(K_{c}=10^{-3}\)
Higher value of Kc means high concentration of products and low concentration of the reactants.
At 700 K, the equilibrium constant for the reaction \(H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)\) is 54.8. If 0.5 mol \(l^{-1}\) of HI(g) is present at equilibrium at 700 K, assuming that we initially started with HI(g) and allowed to reach equilibrium at 700 K. the concemtration of \(H_{2}\) or \(I_{2}\) each would be
- (a)
\(6.56\times10^{-3} mol\ l^{-1}\)
- (b)
\(6.56\times10^{-5} mol\ l^{-1}\)
- (c)
\(6.56\times10^{-6} mol\ l^{-1}\)
- (d)
\(6.56\times10^{-8} mol\ l^{-1}\)
For reaction H2(g)+I2 (g)⇌2HI(g)
Kc=\(\frac{[HI]^{2}}{[H_{2}][I_{2}]}\)=54.8
Let x mol-1 be the concentration of H2(g) and I2(g) at equilibrium, and given (HI)=0.50 mol l-1 at equilibrium
We have 54.8=\(\frac{(0.50mol l^{-1})^{2}}{x\times x}\)
or x2=\(\frac{0.25mol^{2}l^{-2}}{54.8}\)
or x2=\(\frac{0.25 mol^{2}l^{-2}}{54.8}\)
or x2=\(\frac{0.25 mol^{2}l^{-2}}{54.8}\)
or x=6.56x10-3 mol l-1
The equilibrium constant at 298 K for \(Cu(s)+2Ag(aq)\rightleftharpoons Cu^{2}(aq)+2Ag(s)\) is \(2.0 \times 10^{15}\). In a solution, the concentration of \(Cu^{2+}\) ions is \(1.8\times10^{-2} mol \ l^{-1}\), and the concentration of \(Ag^{+} \) ions \(3.0 \times 10^{-19}\) mol \(l^{-1}\). The above system under above condition
- (a)
Will move in forward direction
- (b)
Will move in backward direction
- (c)
will remain in equilibrium
- (d)
it cannot be predicted
For the reaction KL=\(\frac{[Cu^{2+}(aq)}{[Ag^{+}(aq)]^{2}}\)
Given [Cu2+(aq)]=1.8x10-2 mol l-1 and [Ag+(aq)]=3x10-9 mol l-1]
The value of \(\frac{[Cu^{2+}(aq)]}{[Ag^{+}(aq)]^{2}}\)
=\(\frac{1.8\times10^{-2}mol l^{-1}}{(3.0\times10^{-9})^{2}mol^{2}l^{-2}}=2\times10^{15}\)mol l-1
Since this value is equal to the given value of K(=2x1015 ),hence the system is at equilibrium
One mole of A and one mole of B are mixed in one litre container. At equilibrium, the conceration of C was found to be 0.6 mol \(l^{-1}\). K for the reaction \(A(g)+B(g)\rightleftharpoons C(g)+D(g)\) is
- (a)
2.25
- (b)
2.50
- (c)
4.50
- (d)
9.0
As the volume is 1 litre Kc=\(\frac{0.6\times{0.6}}{0.4\times{0.4}}\)=2.25
For the reaction \(A(g)+B(g)\rightleftharpoons C(g)\) the equilibrium partical pressures are \(P_{A}\)=0.15 atm, \(P_{B}\)+0.10 atm and \(P_{c}=0.30 \) atm. The volume was reduced so that reestablishing the equilibrium the partical pressure of A and B were doubled. The partical pressure of C would
- (a)
0.40 atm
- (b)
0.80 atm
- (c)
1.20 atm
- (d)
1.80 atm
Kp=\(\frac{P_{C}}{P_{A}\times{P}_{B}}\)=\(\frac{0.30atm}{0.15atm\times{0.1atm}}\)=20 atm
Now 20=\(\frac{P_{C}}{0.15\times{2}\times{0.1}\times{2}}\)
or PC=1.20 atm
The ester, ethyl acetate, is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as \(CH_{3}COOH(l)+C_{2}H_{5}OH(l)\rightleftharpoons CH_{3}COOC_{2}H_{5}(l)+H_{2}O(l)\) At 293 K, If one starts with 1.000 mole of acetic acid and 0.180 mole of ethanol, there is 0.171 mole of ethyl acetate in the final equilibrium mixture. The equilibrium would be
- (a)
3.919
- (b)
2.919
- (c)
4.919
- (d)
5.919
CH3COOH(l)+C2H5OH(l) ⇌ CH3COOC2H5(l)+H2O(l)
CH3COOH | C2H5OH | CH3COOC2H5 | H2O | |
Initial | 1 mole | 0.18 mole | 0 | 0 |
At Equilibrium | (1-0.171) mol | (0.18-0.171) mole | 0.171 mole | 0.171 mole |
Kc=\(\frac{(0.171/V)(0.171/V)}{(1-0.171)/V\times(0.18-0.171)/V}\)=3.919
Under equilibrium conditions the amount of \(SO_{3}\) in a three-litre vessel was found to be 8g. Its equilibrium concentration could be
- (a)
\(0.8 \ mol \ l^{-1}\)
- (b)
\(0.33 \ mol \ l^{-1}\)
- (c)
\(3.3 \ mol \ l^{-1}\)
- (d)
\(8.8 \ mol \ l^{-1}\)
8g SO3=\(\frac{8}{80}\)moles=\(\frac{1}{10}\)mole=0.1 mole
Molar concentration=\(\frac{moles}{litre}\)=\(\frac{0.1}{3}\)=0.033 mol l-1
The equilibrium constant for the reaction \(2SO_{2}+O_{2}\rightleftharpoons 2SO_{3}\) at a particular temperature is 3.5 \(atm^{-1}\). If equal volumes of SO_{2} and SO_{3} are formed at equalilibrium, the partial pressure of oxygen would be
- (a)
\(0.35 \ atm\)
- (b)
\(0.29 \ atm\)
- (c)
\(0.17 \ atm\)
- (d)
\(0.11 \ atm\)
Kp=\(\frac{P(SO_{3})^{2}}{P(SO_{2})^{2}\times{PO_{2}}}\)
Equal volume means equal moles of SO2 and SO3 , hence the partial pressures of SO2 and SO3 will both be the same in the mixtures, i.e.,
\(P_{{O}_{2}}=\frac{1}{K_{p}}=\frac{1}{3.5}=0.29atm\)
A reaction is \(A+B\rightleftharpoons C+D.\) Initially, we start with equal concentrations of A and B. At equlilibrium we find the moles of C is two times of A. What is the equilibrium constant of the reaction ?
- (a)
2
- (b)
4
- (c)
1/2
- (d)
1/4
A+B ⇌ C+D
A | B | C | D | |
Initial | 1 | 1 | 0 | 0 |
At Equilibrium | 1-∝ | 1-∝ | ∝ | ∝ |
According to the given condition at equilibrium ∝=2(1-∝) or ∝=2/3
K=\(\frac{\alpha^{2}}{(1-\alpha)^{2}}=\frac{\left(\frac{2}{3}\right)^{2}}{\left(\frac{1}{3}\right)^{2}}\)=4
If \(K_{p}\) for a reaction \(A(g)+2B(g)A+B\rightleftharpoons 3C(g)+D(g)\) is 0.05 atm at 1000 K, its \(K_{c}\) in terms of R will be
- (a)
2,0000 R
- (b)
0.02 R
- (c)
\(5\times10^{-15}\) R
- (d)
\(5\times10^{-5} R^{-1}\)
Kp=Kc(RT)∆n
Here Δn=4-3=1
ஃ Kc=\(\frac{K_{p}}{RT}=\frac{0.05}{R\times100}\)=5x10-5R-1
The equlilibrium constant for the gaseous reaction A+B\(??\rightleftharpoons \)C+D is 100 at \(25^{o}\)C. Consider the following statement in this regard. If the initial concentrations of all the four species were 1.0 M, then the equilibrium concentration of
(1) A would be 0.182 mol and
(2) C would be 0.818 mol and
(3) D would be 1.818 mole
Of the statements
- (a)
1 and 2 are correct
- (b)
1 and 3 are correct
- (c)
2 and 3 are correct
- (d)
1,2 and 3 are correct
A+B⇌C+D
A | B | C | D | |
Initial | 1 | 1 | 1 | 1 |
At Equilibrium | 1-∝ | 1-∝ | 1+∝ | 1+∝ |
100=\(\frac{(1+\alpha)^{2}}{(1-\alpha)^{2}}\) or 10=\(\frac{1+\alpha}{1-\alpha}\)
or ∝=0.818
A sample of pure \(PC1_{3}\) was introduces into an evacuated vessel at 473 K. After equilibrium was attained, concentration of \(PC1_{3}\) was found to be \(0.5\times 10^{-1} mol L^{-1}\). The value of \(K_{c}\) is \(8.3\times10^{-13}\), the concentration of \(PC1_{3}\) at equilibrium would be
- (a)
0.01 \(mol L^{-1}\)
- (b)
0.02 \(mol L^{-1}\)
- (c)
0.03 \(mol L^{-1}\)
- (d)
0.04 \(mol L^{-1}\)
PCl5 \(\rightleftharpoons\) PCl3 + Cl2
At equi. 0.5 X 10-1 mol L-1 x x
Kc = \({ { [{PCl}_{3}][{Cl}_{2}] }\over{ [{PCl}_{5}] } }\)
or 8.3 X 10-3 = \({ { [x][x] }\over{ 0.5\times{10}^{-1}} }\)
or x2 = 4 X 10-4
x = 2 X 10-2
= 0.02 mol L-1
The equilibrium constant in a reversible reaction at specified temperature
- (a)
does not depend on the initial concentrations
- (b)
depend on the initial concentrations of the reactants
- (c)
depend on the concentrations of the products at equilibrium
- (d)
is not characteristic of the reaction
The equilibrium constant in a reversible reaction at a specified temperature does not depend on the initial concentration.
The unit of equilibrium constant, K for the reaction, \(A+B\rightleftharpoons C\) would be
- (a)
mol L-1
- (b)
mol L
- (c)
mol-1 L
- (d)
\(1\over mol \ L\)
\(A+B\rightleftharpoons C\)
\(K=\frac { \begin{bmatrix} C \end{bmatrix} }{ \begin{bmatrix} A \end{bmatrix}\begin{bmatrix} B \end{bmatrix} } =\frac { mol\quad L^{ -1 } }{ (mol\quad { L }^{ -1 })(mol\quad { L }^{ -1 }) } \)
=(mol L-1)-1=mol-1L
For the reaction Fe(s)+S(s)\(\rightleftharpoons \)FeS(s); the expression for equilibrium constant is
- (a)
\(\frac { \begin{bmatrix} FeS \end{bmatrix} }{ \begin{bmatrix} Fe \end{bmatrix}\begin{bmatrix} S \end{bmatrix} } \)
- (b)
\(\frac { { \begin{bmatrix} Fe \end{bmatrix}\begin{bmatrix} S \end{bmatrix} } }{ { \begin{bmatrix} FeS \end{bmatrix} } } \)
- (c)
\(\begin{bmatrix} Fe \end{bmatrix}\begin{bmatrix} S \end{bmatrix}\begin{bmatrix} FeS \end{bmatrix}\)
- (d)
None of these
The expression for equilibrium constant is not represented by these equations because concentration of reactants in pure state is zero.
For the reaction, \(PCl_{ 3 }(g)+Cl_{ 2 }(g)\rightleftharpoons PCL_{ 5 }(g)\) the position of equilibrium can be shifted to the right side by
- (a)
doubling the volume
- (b)
increasing the temperature
- (c)
addition of Cl2 at constant volume
- (d)
addition of equimolar quantities of PCl3 and PCl5
For the reaction, \(PCl_{ 3 }(g)+Cl_{ 2 }(g)\rightleftharpoons PCl_{ 5 }(g)\) addition of Cl2 at constant volume shifted the reaction to the right side.
What is the effect of having the pressure by doubling the volume on the following reaction at 500oC?
\(H_{ 2 }(g)+I_{ 2 }(g)\rightleftharpoons 2HI(g)\)
- (a)
Shift to products side
- (b)
Shift to reactants side
- (c)
Liquefaction of HI
- (d)
No effect
We know that, pV=nRT
p become \({1\over 2}\) p and V becomes 2V, so
\({1\over 2}pX2V=pV=nRT\)
Hence, there is no effect upon reaction.
In which of the following reactions, increases in the volume at constant temperature do not affect the number of moles at equilibrium?
- (a)
\(2NH_{ 3 }\rightleftharpoons N_{ 2 }+3H_{ 2 }\)
- (b)
\(C(s)+\frac { 1 }{ 2 } O_{ 2 }(g)\longrightarrow CO(g)\)
- (c)
\(H_{ 2 }(g)+O_{ 2 }(g)\longrightarrow H_{ 2 }O_{ 2 }(g)\)
- (d)
None of the above
In these reaction, equilibrium, \(\Delta G\)=0under conditions of constant temperature and pressure.
At constant temperature, the equilibrium constant (Kp) for the decomposition reaction \(N_{ 2 }O_{ 4 }(g)\rightleftharpoons 2NO_{ 2 }(g)\), is expressed by \(K_{ p }=\frac { (4x^{ 2 }p) }{ (1-x^{ 2 }) } \) where, p=pressure, x=extent of decomposition. Which one of the following statements is correct?
- (a)
Kp increases with increase of p
- (b)
Kp remains constant with change in p and x
- (c)
Kp increases with increase of x
- (d)
None of the above
Kp remains constant with change in p and x
For a system in equilibrium, \(\Delta G=0\) under conditions of constant
- (a)
temperature and pressure
- (b)
energy and volume
- (c)
temperature and energy
- (d)
pressure and volume
For a system in equilibrium, \(\Delta G=0\) under conditions of constant temperature and pressure.
Ice and water are in equilibrium at 273K, which of the following statement is correct?
- (a)
\(G_{(ice)} > G_{({H_2O})}\)
- (b)
\(G_{(ice)}< G_{(H_2O)}\)
- (c)
\(G_{(ice)}=G_{(H_2O)}=0\)
- (d)
\(G_{(ice)}=G_{(H_2O)}\ne0\)
At equilibrium \(\Delta G=0\)
i.e., \(G_{(ice)}=G_{(H_2O)}\ne 0\)
In the thermal decomposition of potassium chlorate given as \(2KClO_{ 3 }\longrightarrow 2KCl+3O_2\); Law of mass action
- (a)
Can be applied
- (b)
cannot be applied
- (c)
can be applied at low temperature and pressure
- (d)
can be applied at high temperature and pressure
Thermal decomposition of potassium chlorate is irreversible. So, law of mass action cannot be applied on this reaction.
The decomposition of N2O4 to NO2 is carried out at 280K in chloroform. When equilibrium has been established, 0.2 mole of N2O4 and 2X10-3 mole of NO2 are present in 2L solution. The equilibrium constant for reaction, \(N_{ 2 }O_{ 4 }\rightleftharpoons 2NO_{ 2 }\) is
- (a)
1X10-6
- (b)
1X10-3
- (c)
1X10-4
- (d)
1X10-5
N2O4 \(\rightleftharpoons\) 2NO2
Kc=\({ { {[{NO}_{2}]}^{2} }\over{[{N}_{2}{O}_{4}] } }={ { \left[ 2\times{ { {10}^{-3} }\over{2 } } \right]^{2} }\over{\left[ { { 0.2 }\over{ 2} } \right] } }={ { {10}^{-6} }\over{{10}^{-1} } }=1\times{10}^{-5}\)
Which of the following options will be correct for the stage of half completion of the reaction \(A\rightleftharpoons B\)?
- (a)
\(\Delta G^o=0\)
- (b)
\(\Delta G^o>0\)
- (c)
\(\Delta G^o<0\)
- (d)
\(\Delta G^o=-RT\ ln \ 2\)
\(\Delta G^o\)=-2.303RT log K
At the stage, of half completion of reaction \(\begin{bmatrix} A \end{bmatrix}=\begin{bmatrix} B \end{bmatrix}\)
Therefore, K=1
Thus, \(\Delta G^o\)=0
In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?
- (a)
\(H_{ 2 }(g)+I_{ 2 }(g)\rightleftharpoons 2HI(g)\)
- (b)
\(PCl_{ 5 }(g)\rightleftharpoons 2PCl_{ 3 }(g)+Cl_{ 2 }(g)\)
- (c)
\(N_{ 2 }(g)+3H_{ 2 }(g)\rightleftharpoons 2NH_{ 3 }(g)\)
- (d)
The equilibrium will remain uneffected in all the three cases
The equilibrium will remain unaffected in all the three cases on the addition of small amount of argon at constant volume.
One mole of nitrogen and three moles of hydrogen are mixed in a litre container. If 0.25 percent of nitrogen is converted to ammonia by the following reaction \(N_{ 2 }(g)+3H_{ 2 }(g)\rightleftharpoons 2NH_{ 3 }(g)\) Calculate the equilibrium constant (Kc) in concentration units.
- (a)
1.49X10-5 L2mol-2
- (b)
1.49X105L2 mol-2
- (c)
1.19X10-5 L2 mol-2
- (d)
1.19X105L2mol-2
N2(g)+2H2(g)\(\rightleftharpoons\) 2NH3
initial | 1 | 3 | 0 |
At equilibrium | (1-x) | (3-3x) | 2x[ \(\because\) x = 0.0025 ] |
Active mass | \({ { (1-0.0025) }\over{ 4} }\) | \({ { 3-0.0075 }\over{ 4 } }\) | \({ { 0.0050 }\over{ 4 } }\) |
Applying law of motion action,
Kc=\({ { {[{NH}_{3}]}^{2} }\over{ [{N}_{2}]} {[{N}_{2}]}^{3}}\)
= \({ { {\left( { { 0.0050 }\over{4 } } \right)}^{2} }\over{\left( { {0.9975 }\over{ 4} } \right) \left( { {2.9925 }\over{4 } } \right)^{3}} }\)
= 1.49 X 10-5 L2 mol-2
The partial pressure of B is found to be one fourth of the partial pressure of A. For a reaction, \(A(g)\rightleftharpoons B(g)\). The value of \(\Delta G^o\) of the reaction \(A\rightarrow B\) will be
- (a)
RT ln 4
- (b)
-RT ln 4
- (c)
2.303RT ln 4
- (d)
-2.303RT ln 4
\(K^o_{eq}={1\over 4}\)
\(\Delta G^o=-RT \ ln \ 1/4=RT \ ln \ 4\)
The equilibrium constant (Kc) for the reaction, \(N_{ 2 }(g)+O_{ 2 }(g)\longrightarrow 2NO(g)\), at temperature T is 4X10-4. The value of Kc for the reaction.
\(NO(g)\longrightarrow \frac { 1 }{ 2 } N_{ 2 }(g)+\frac { 1 }{ 2 } O_{ 2 }(g)\) at the same temperature is
- (a)
0.02
- (b)
2.5X102
- (c)
4X10-4
- (d)
50.0
N2(g) + O2(g) \(\longrightarrow\) 2NO(g)
Kc = \({ { {[NO]}^{2} }\over{[{N}_{2}][{O}_{2}] } }=4\times{10}^{-4}\)
2NO(g) \(\longrightarrow\) N2(g)+O2(g)
\({K'}_{c}={ { 1 }\over{ {K}_{c} } }={ { [{N}_{2}][{O}_{2}] }\over{{[NO]}^{2} } }={ { 1 }\over{ 4\times{10}^{-4} } }={ {{10}^{4} }\over{ 4 } }\)
NO(g) \(\longrightarrow\) \({ { 1 }\over{ 2 } }\)N2(g) + \({ { 1 }\over{ 2 } }\)O2(g)
\({K''}_{c}={ { {[{N}_{2}]}^{1/2}{[{O}_{2}]}^{1/2} }\over{ [NO] } }=\sqrt{{K'}_{c}}=\sqrt{{ {{10}^{4} }\over{ 4 } }}={ { 100 }\over{ 2 } }= 50\)
The exothermic formation of CIF3 is represented by the equation \(Cl_2(g)+3F_2(g)\rightleftharpoons 2CIF_3(g);\Delta H_r=-329kJ\) which of the following will increase the quantity of CIF3 in an equilibrium mixture of Cl2, F2 and CIF3?
- (a)
Adding F2
- (b)
Increasing the volume of the container
- (c)
Removing Cl2
- (d)
Increasing the temperature
Reaction is exothermic. By Le-Chatelier principle, a reaction is spontaneous in forward side (in the direction of formation of more CIF3), if F2 is added, temperature is lowered and CIF3 is removed.