Equilibrium Test

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Equilibrium Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

For the gas phase reaction : $2NO(g) \leftrightharpoons N_2(g) +O_2 (g) ; \Delta H = -43.5$ kcal. Which one of the following is true for the reaction: $N _2(g) + O_2(g) \leftrightharpoons 2NO(g)$ ?

A
K is independent of T

B
K decreases as T decreases

C
K increases as T decreases

D
K varies with with addition of NO

Solution

Le Chatelier's Principle is applied for both the reactions to predict the direction of equilibrium.

For equation,

$2NO\rightleftharpoons N_2+O_2$

The above reaction is exothermic as the value of enthalpy is negative. The value of $K$ decreases with increase in temperature. The equilibrium will shift to the left to compensate the increase in temperature.

For reaction,

$N_2+O_2\rightleftharpoons 2NO$

For the above equation value of enthalpy will be same $(+43.5\space Kcal)$ but with positive. It is an endothermic reaction. A decrease in temperature will lead to a decrease of $K$ as equilibrium will shift to the left.

Hence, the correct option is (B).
Question 2
1 / -0

If $K_1$ and $K_2$ are the equilibrium constants for a reversible reaction at $T_1 K$ and $T_2 K$ temperature, respectively $( T_1 < T_2 )$ and the reaction takes place with neither heat evolution nor absorption, then

A
$K_1 > K_2$ at high temperature

B
$K_1 < K_2$ at high temperature

C
$K_1 = K_2$ only at high temperature

D
$K_1 = K_2$ at any temperature

Solution

We know that,
The van't hoff equation gives relation between equilibrium constant and temperature when enthalpy is given.

$log\frac{K_2}{K_1}=\frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

For given reversible reaction, $K_1$ and $K_2$ are the equilibrium constants at temperature $T_1$ and $T_2$ . And $R$ is gas constant.

Heat is neither evolved nor absorbed. Therefore, enthalpy change, $\Delta H=0$ .

When $\Delta H=0$ and $T_1<T_2$ ,

$log\frac{K_2}{K_1}=\frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]=0$

Therefore, $K_2/K_1=1$ or $K_1=K_2$ .

Hence, the correct option is (D).
Question 3
1 / -0

In a closed container following equilibrium will be attained:
$A(s)+B(g) \leftrightharpoons AB(g)$
$B(g)+C(g) \leftrightharpoons BC(g)$
On adding $He$ gas (inert) to the above system at constant pressure & temperature:

A

The amount of AB(g) will be increased surely

B
The amount of B(g) will be decreased surely

C
The amount of BC(g) will be decreased surely.

D
The amount of C(g) will be decreased surely.
Question 4
1 / -0

n-caproic acid, $C_{5}H_{11}COOH$ , found in coconut and palm oil is used in making artificial flavors, has solubility in water equal to $11.6 g/L$ . The saturated solution has $pH = 3.0$ . The $K_{a }$ of acid is

A
$10^{-6}$

B
$10^{-5}$

C
$2 \times 10^{-5}$

D
$2 \times 10^{-6}$
Question 5
1 / -0

$O_3$ is prepared by subjecting $O_2$ to silent electric discharge. The favourable conditions for the formation of ozone according to Le-chatlier's principle are

A
low temperature, low pressure

B
high temperature, high pressure

C
low temperature, high pressure

D
high temperature, low pressure

Solution
Question 6
1 / -0

A solution has initially $0.1 M -HCOOH$ and $0.2 M- HCN$ . $K_{a}$ of $HCOOH = 2.56 \times 10^{-4}$ , $K_{a}$ of $HCN = 9.6 \times 10^{-10}$ . The only incorrect statement for the solution is $(log 2 = 0.3)$

A
$[H]^{+} = 1.6 \times 10^{-3} M$

B
$[HCOO]^{-} = 1.6 \times 10^{-3} M$

C
$[CN]^{-} = 1.2 \times 10^{-7} M$

D
$pOH = 2.8$
Question 7
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Small amount of freshly precipitated magnesium hydroxides are stirred vigorously in a buffer solution containing $0.25 M$ of $NH_{4}Cl$ and $0.05 M$ of $NH_{4}OH$ . $[Mg^{2+}]$ in the resulting solution is ( $K_{b}$ for $NH_{4}OH = 2.0 \times 10^{-5}$ and $K_{sp}$ of $Mg(OH)_{2} = 8.0 \times 10^{-12}$ )

A
$4 \times 10^{-6} M$

B
$2 \times 10^{-6} M$

C
$0.5 M$

D
$2.0 M$

Solution
Question 8
1 / -0

In a system : $A(s) \leftrightharpoons 2B(g) +3C(g)$ if the concentration of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of B to decrease by

A
two times the original value

B
one half of its original value

C
$2 \sqrt 2$ times its original value

D
8 times its original value

Solution

Sol.
The given reaction is :

$A(s) \rightleftharpoons 2B(g) + 3C(g)$

The equilibrium constant will remain the same if the concentration of C is increased by a factor of 2.

Let the concentration of A, B and C be x, y and z.

So, we can say that

$K_{c}=[C]^3[B]^2=(z)^{3}(y)^{2}$

Now, the new concentration of C is 2z, so let new concentration of B be $y^{'}$ .

$K_{c} =(y^{'})^{2}(2z)^{3}$

As the equilibrium constant doesn't change during equilibrium.

$(z)^{3}(y)^{2}$ $=(y^{'})^{2}(2z)^{3}$

On solving, we get

$\Rightarrow y^{'} =\dfrac{y}{2\sqrt{2} }$

Hence, option (C) is correct.
Question 9
1 / -0

The reaction : $MgCO_3(s) \leftrightharpoons MgO(s) + CO_2(g)$ is in progress. if number of mole of MgO in the vessel is doubled at an instance?

A
the reaction quotient , Q is havled

B
the reaction quotient , Q is doubled

C
the moles of $CO_2$ present at equilibrium is havled

D
the partial pressure of $CO_2$ in the vessel remains unchanged

Solution
Question 10
1 / -0

The EMF of cell: $H_{2}(g)|$ Buffer $\parallel$ Normal calmelelectrod, is 0.70 V at $25^{\circ}C,$ when barometric pressure is 760 mm. What is the pH of the buffer solution? $E_{Calomel}^{0} = 0.28 V. [2.303 RTIF = 0.06]$

A
3.5

B
7.0

C
tending to zero

D
tending to 14.0

Solution

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Equilibrium Test - 70

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Equilibrium Test - 70

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Question 1

K is independent of T

K decreases as T decreases

K increases as T decreases

K varies with with addition of NO

Solution

Question 2

K1>K2 K_1 > K_2 K1​>K2​ at high temperature

K1<K2 K_1 < K_2 K1​<K2​ at high temperature

K1=K2 K_1 = K_2 K1​=K2​ only at high temperature

K1=K2 K_1 = K_2 K1​=K2​ at any temperature

Solution

Question 3

The amount of AB(g) will be increased surely

The amount of B(g) will be decreased surely

The amount of BC(g) will be decreased surely.

The amount of C(g) will be decreased surely.

Question 4

10−610^{-6}10−6

10−510^{-5}10−5

2×10−52 \times 10^{-5}2×10−5

2×10−62 \times 10^{-6}2×10−6

Question 5

low temperature, low pressure

high temperature, high pressure

low temperature, high pressure

high temperature, low pressure

Solution

Question 6

[H]+=1.6×10−3M[H]^{+} = 1.6 \times 10^{-3} M[H]+=1.6×10−3M

[HCOO]−=1.6×10−3M[HCOO]^{-} = 1.6 \times 10^{-3} M[HCOO]−=1.6×10−3M

[CN]−=1.2×10−7M[CN]^{-} = 1.2 \times 10^{-7} M[CN]−=1.2×10−7M

pOH=2.8pOH = 2.8pOH=2.8

Question 7

4×10−6M4 \times 10^{-6} M4×10−6M

2×10−6M2 \times 10^{-6} M2×10−6M

0.5M0.5 M0.5M

2.0M2.0 M2.0M

Solution

Question 8

two times the original value

one half of its original value

22 2 \sqrt 2 22​ times its original value

8 times its original value

Solution

Question 9

the reaction quotient , Q is havled

the reaction quotient , Q is doubled

the moles of CO2 CO_2 CO2​ present at equilibrium is havled

the partial pressure of CO2 CO_2 CO2​ in the vessel remains unchanged

Solution

Question 10

3.5

7.0

tending to zero

tending to 14.0

Solution

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$K_1 > K_2$ at high temperature

$K_1 < K_2$ at high temperature

$K_1 = K_2$ only at high temperature

$K_1 = K_2$ at any temperature

$10^{-6}$

$10^{-5}$

$2 \times 10^{-5}$

$2 \times 10^{-6}$

$[H]^{+} = 1.6 \times 10^{-3} M$

$[HCOO]^{-} = 1.6 \times 10^{-3} M$

$[CN]^{-} = 1.2 \times 10^{-7} M$

$pOH = 2.8$

$4 \times 10^{-6} M$

$2 \times 10^{-6} M$

$0.5 M$

$2.0 M$

$2 \sqrt 2$ times its original value

the moles of $CO_2$ present at equilibrium is havled

the partial pressure of $CO_2$ in the vessel remains unchanged