Tag: equilibrium in chemical processes

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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

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

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

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

0.1 mole of $N _2O _4(g)$ was sealed in a tube under one atmospheric conditions at $25^0C$. Calculate the number of moles of $NO _2(g)$ present, if the equilibrium $N _2O _4(g)\rightleftharpoons  2NO _2(g)$ $(K _p=0.14)$ is reached after some time.

  1. $1.8 \times 10^2$
  2. $2.8 \times 10^2$

  3. 0.034
  4. $2.8 \times 10^{-2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to the equation

$\begin{matrix}  & N _2O _4& \rightleftharpoons & 2NO _2 \ \text{Initial}&1 \, atm  & &0 \ \text{change}& -x&  &+2x\ \text{Equilibrium}&1-x&&2x  \end{matrix}$
$K _P = \dfrac{(P _{NO _2})^2}{P _{N _2O _4}}$   $[\therefore K _P$ is similar to $K _C$ in aspect of setting up rate quotient]

$0.14 = \dfrac{(2x)^2}{1-x}$

$0.14(1-x) = 4x^2$
$x = 0.17$
From ideal gas equation
$PV = nRT$

$V = \dfrac{nRT}{P}$

$V = \dfrac{0.1\times 0.082 \times (273 + 25)}{1}$

$V = 2.45 lit$

Moiles of $NO _2= \dfrac{P _{NO _2} \times V}{RT}$

Moles of $NO _2 = \dfrac{0.34\times 2.45}{0.0821 \times (273 + 25)}$

Moles of $NO _2 = 0.034$

$\therefore$ option C is correct

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

Consider the following reactions in which all the reactants and the products are in 


$2PQ \rightleftharpoons P _2 + Q _2 ; K _1 = 2.5 \times 10^5$

$PQ + \cfrac{1}{2} R _2 \rightleftharpoons PQR; K _2 = 5 \times 10^{-3}$

The value of $K _3$ for the equilibrium $\cfrac{1}{2} P _2 + \cfrac{1}{2} Q _2 + \cfrac{1}{2} R _2 \rightleftharpoons PQR,$ is:

  1. $2.5 \times 10^{-3}$
  2. $2.5 \times 10^{3}$
  3. $1.0 \times 10^{-5}$
  4. $5 \times 10^{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To obtain the target equilibrium equation, multiply the first reaction by -1/2 (or reverse and take square root) and add the second reaction. This corresponds to combining the equilibrium constants as K3 = (1 / sqrt(K1)) * K2.

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

Chemical equilibria is (I) _________ in nature. It occurs in (2) ________ reactions only.  At equilibria, all (3) __________ properties becomes constant. Chemical equilibrium gets affected by change in temperature, pressure, concentration, volume etc but is not altered by addition of (4)________. Equilibria are of two types (5) ________ and homogeneous equilibria.

  1. (1) dynamic (2) reversible (3) observable (4) catalyst (5) homogeneous

  2. (1) static (2) irreversible (3) observable (4) reactants (5) heterogeneous

  3. (1) dynamic (2) irreversible (3) observable (4) reactants (5) homogeneous

  4. (1) dynamic (2) reversible (3) observable (4) catalyst (5) heterogeneous

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Chemical equilibria is dynamic in nature. It occurs in reversible reactions only. At equilibria all observable properties becomes constant. Chemical equilibrium is affected by a change in temperature, pressure, concentration, volume etc but is not altered by the addition of catalyst. Equilibria is of two types heterogeneous and homogeneous equilibria.

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

Which of the following statements is correct?

  1. In equilibrium mixture of ice and water kept in perfectly insulated flask, mass of ice and water does not change with time.

  2. The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.

  3. On addition of catalyst, the equilibrium constant value is not affected.
  4. Equilibrium constant for a reaction with negative $\triangle$H value decreases as the temperature increases.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equilibrium reaction containing $Fe(III)$ nitrate & $KSCN$ is given as :-

${ Fe }^{ 3+ }\left( aq \right) +{ SCN }^{ - }\left( aq \right) \rightleftharpoons { \left[ Fe\left( SCN \right)  \right]  }^{ 2+ }\left( aq \right) $
Here, the red colour in the reactions occurs due to formation of ${ \left[ Fe\left( SCN \right)  \right]  }^{ 2+ }$. Now, when oxalic acid is added, it will react with ${ Fe }^{ 3+ }$ to form ${ \left[ Fe{ \left( { C } _{ 2 }{ O } _{ 4 } \right)  } _{ 3 } \right]  }^{ 3- }$. So, the concentration of ${ Fe }^{ 3+ }$ ions will get decreased. By applying Lechatelier's principle we see that the reaction will proceed towards backward direction and concentration of ${ \left[ Fe\left( SCN \right)  \right]  }^{ 2+ }$ will get decreased. Consequently the intensity of red colour is decreased.