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Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases
A reaction continues even after the attainment of equilibrium.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A reaction continues even after the attainment of equilibrium.
The reaction proceeds in forward as well as reverse direction. When equilibrium is attained, it is dynamic in nature. The reactants are converted into products through forward reaction and the products are converted into reactants through reverse reaction.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases
The equilibrium state can be attained from both sides of the chemical reaction.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equilibrium state can be attained from both sides of the chemical reaction.
Thus, if initially, the concentration of reactants is much higher than the equilibrium concentration and the concentration of products is much lower than the equilibrium concentration, the reaction will proceed in the forward direction till equilibrium is attained. On the other hand, if initially, the concentration of reactants is much lower than the equilibrium concentration and the concentration of products is much higher than the equilibrium concentration, the reaction will proceed in the reverse direction till equilibrium is attained.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

For hypothetical reversible reaction $\dfrac {1}{2}A _{2}(g) + \dfrac {1}{2}B _{2}(g) \rightarrow AB _{3}(g); \triangle H = -20\ KJ$ if standard entropies of $A _{2}, B _{2}$ and $AB _{3}$ are $60, 40$ and $50\ JK^{-1} mole^{-1}$ respectively. The above reaction will be in equilibrium at

  1. $400\ K$
  2. $500\ K$
  3. $250\ K$
  4. $200\ K$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

The equilibrium constant for a reaction is $K$, and the reaction quotient is $Q$. For a reaction mixture, the ratio $\dfrac {K}{Q}$ is $0.33$. This means that:

  1. the reaction mixture will equilibrium to form more reactant species

  2. the reaction mixture will equilibrium to form more product species
  3. the equilibrium ratio of reactant to product concentrations will be $3$
  4. the equilibrium ratio of reactant to product concentrations will be $0.33$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that for a reaction if $\dfrac{K}{Q} < 1,$ the reaction proceeds in backward direction. Hence, the reaction mixture forms more reactant species.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

Assume that the decomposition of $H{ NO } _{ 3 }$ can be represented by the following equation
$4H{ NO } _{ 3 }(g)\rightleftharpoons 4{ NO } _{ 2 }(g)+2{ H } _{ 2 }O(g)+{ O } _{ 2 }(g)\quad $'and the reaction approaches equilibrium at $400K$ temperature and $30$ atm pressure. The equilibrium partial pressure of $H{ NO } _{ 3 }$ is $2$ atm
Calculate ${K} _{c}$ in ${ \left( mol/L \right)  }^{ 3 }$
(Use: $R=0.08atm-L/mol-K$)

  1. $4$
  2. $8$
  3. $16$
  4. $32$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the relation between Kp and Kc, namely Kp = Kc * (RT)^delta_n, we find delta_n = (4 + 2 + 1) - 4 = 3. Given P(HNO3) = 2 atm and total pressure 30 atm, we can find partial pressures at equilibrium, then compute Kp and subsequently Kc.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

The optical rotation of the $\alpha-form$ of a pyramose is $+150.7^{\circ}$, that of the $\beta - form$ is $+52.8^{\circ}$. In solution an equilibrium mixture of these anomers has an optical rotation of $+80.2^{\circ}$. The percentage of the $\alpha$ form in equilibrium mixture is:

  1. $28$%
  2. $32$%
  3. $68$%
  4. $72$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\alpha $ from $=+150.{ 7 }^{ 0 }$, $\beta $ from $=+52.{ 8 }^{ 0 }$

at equilibrium optical rotation $=+80.2$ 
let, at equilibrium $\alpha $-from exist $=x$
      at equilibrium $\beta $-from exist $=(100-x)$
Therefore, $\dfrac { 150.7x+\left( 100-x \right) \times 52.8 }{ 100 } =80.2$
$\Rightarrow \quad 150.7x+5280-52.8x=8020$
$\Rightarrow \quad 99.9x=2740$
$\Rightarrow \quad x=27.42\approx 28$%

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

A reaction mixture containing $H _{2}, N _{2}$ and $NH _{3}$ has partial pressure $2\ atm, 1\ atm$ and $3\ atm$ respectively at $725\ K$. If the value of $K _{P}$ for reaction, $N _{2} + 3H _{2}\rightleftharpoons 2NH _{3}$ is $4.28\times 10^{-5} atm^{-2}$ at $725\ K$, in which direction the net reaction will go :

  1. Forward

  2. Backward

  3. No net reaction

  4. Direction of reaction cannot be predicted
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

                ${ N } _{ 2 }+{ 3H } _{ 2 }\rightleftharpoons 2{ NH } _{ 3 }$

at eqn,   1atm    2atm       3atm
given, ${ K } _{ p }=4.28\times { 10 }^{ -5 }{ atm }^{ -2 }$
${ Q } _{ p }=\dfrac { { \left( { P } _{ { NH } _{ 3 } } \right)  }^{ 2 } }{ \left( { P } _{ { N } _{ 2 } } \right) { \left( { P } _{ { H } _{ 2 } } \right)  }^{ 3 } } =\dfrac { { \left( 3 \right)  }^{ 2 } }{ 1\times { \left( 2 \right)  }^{ 3 } } =\dfrac { 3\times 3 }{ 2\times 2\times 2 } =\dfrac { 9 }{ 8 } =1.125{ atm }^{ -2 }$
Since, $\boxed { { Q } _{ p }>>{ K } _{ p } } $
Reaction will move in forward direction.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

Which of the following is true :

  1. $pk _{b}$ for $OH^{-}$ is -1.74 at $25^{o}$C
  2. The equilibrium constant for the reaction between HA ($pk _{a} = 4$) and NaOH at $25^{o}$C will be equal to $10^{10}$
  3. The pH of a solution containing 0.1 M HCOOH ($k _{a} = 1.8 \times 10^{-4}$) and 0.1 M HOCN ($k _{a} = 3.2 \times 10^{-4}$) will be nearly (3 -log 7)

  4. All of the above are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

A mixture of three gases P (density 0.90), Q (density 0.178) and R (density 0.42) is enclosed in a vessel at the constant temperature. When the equilibrium is established:

  1. the gas P will be at the top of the vessel

  2. the gas Q will be at the top of the vessel

  3. the gas R will be at the top of the vessel

  4. the gases will mix homogeneously throughout the vessel.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Independent of density of gases, in equilibrium gases will mix homogenously, this comes from the fact that gases occupies entire volume of container. This also can be thought as gases tries to reduce energy. Hence they separate as far as possible.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

The equilibrium constant $K _{c}$ for the reaction $P _{4}(g) \rightleftharpoons 2P _{2}(g)$
is $1.4$ at $400^{\circ}C$. Suppose that $3$ moles of $P _{4}(g)$ and $2$ moles of $P _{2}(g)$ are mixed in $2$ litre container at $400^{\circ}C$. What is the value of reaction quotient $(Q _{c})$?

  1. $\dfrac {3}{2}$
  2. $\dfrac {2}{3}$
  3. $1$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Q _c = \dfrac{[P _2(g)]^4}{[P _4 (g)]}$

= $\dfrac{(1)^2}{(3/2)}$ 
= $\dfrac{2}{3}$