Equilibrium Test

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Equilibrium Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

At $1000^0 C$ and a pressure of 16 atm, the equilibrium constant of the reaction :
$CO_2(g) + C(s) \leftrightharpoons 2CO(g)$
is such that for every nine moles of CO, there is one mole of $CO_2$ for what pressure of the mixture, is the ratio $CO : CO_2 = 4 :1$ ? the temperature remains $1000^0$ C

A
$40.5$ atm

B
$81$ atm

C
$33.75$ atm

D
$6.7$ atm

Solution
Question 2
1 / -0

For the reaction : $2NOBr(g) \leftrightharpoons 2NO(g) +Br_2(g)$ the ratio $\dfrac {K_p}{P}$ , where P is the total pressure of gases at equilibrium and $P_{Br_2} = \dfrac {P}{9}$ at a certain temperature, is

A
$\dfrac {1}{9}$

B
$\dfrac {1}{81}$

C
$\dfrac {1}{27}$

D
$\dfrac {1}{3}$

Solution
Question 3
1 / -0

In the equilibrium mixture: $H_2(g) +I_2(g) \leftrightharpoons 2HI(g)$ , the mole ratios of gases are $\sqrt 2 : \sqrt 2 : 10$ , respectively. What would be the effect on the mole ratio on adding 5 mole of inert gas at constant pressure?

A
no change

B
the new molar ratio becomes $1 : 1 :5$

C
the new molar ratio becomes $2 : 2 : 5$

D
the new molar ratio becomes $2 : 2 : 5\sqrt 2$

Solution

$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

As we know When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in the number of moles of gases.

But in the above equation number of moles on both sides of the equation is the same and that is 2.

Hence, there will no change in the mole ratio.
Question 4
1 / -0

The equilibrium : $SOCl_2(g) \leftrightarrow SO_2(g) +Cl_2(g)$ is attained at $25^0 C$ in a closed container and helium gas is introduced. Which of the following statements is correct?

A
concentration of $SO_2$ is increased

B
more $Cl_2$ is formed

C
concentration of all change

D
concentration will not change

Solution

According to Le Chatelier's Principle, if a system in equilibrium is subjected to a change of concentration, pressure or temperature, the equilibrium shifts in the direction that tends to undo the effect of the change.

In the given reaction the equilibrium is attained in a closed container and helium gas is introduced. When inert gas helium is added to the equilibrium system at constant volume, it will cause the increase in the total pressure of the system. But the partial pressure of each of the reactant as well as product species will not be affected and will remain the same. Therefore, there will be no effect on the concentration and equilibrium.

Hence, the correct option is (D).
Question 5
1 / -0

When pressure is applied to the equilibrium system: Ice $\leftrightharpoons$ water, which of the following phenomenon will happen?

A
more ice will be formed

B
ice will sublime

C
more water will be formed

D
equilibrium will not disturb

Solution

According to Le Chatelier's Principle, if a system in equilibrium is subjected to a change of concentration, pressure or temperature, the equilibrium shifts in the direction that tends to undo the effect of the change.

When the pressure is applied to the equilibrium system between ice and water, the phase transition temperature of ice reduces. Thus the equilibrium shifts more to right side. Also, when pressure is increased the equilibrium shifts in the direction of less volume.
Therefore, more ice will melt into water.

Hence, the correct option is (c).
Question 6
1 / -0

Which of the following hypothetical reactions is favoured by increases of temperature as well as pressure?

A
$A(s) +2B(g) \leftrightharpoons C(g) +D(g); \Delta H = -ve$

B
$A(g) +2B(g) \leftrightharpoons 2C(g) +D(g); \Delta H = +ve$

C
$2A(g) +B(g) \leftrightharpoons 2C(g) +D(s); \Delta H = +ve$

D
$2A(s) +2B(g) \leftrightharpoons 2C(g) +2D(g); \Delta H = -ve$

Solution

Sol.
Increase in heat favors reactions which are endothermic i.e. $\Delta H =positive$ . Also, an increase in pressure favors the reaction to the side with less number of gaseous moles i.e. $\sum n_p< \sum n_r.$

Both the conditions are satisfied in (C) option.

Hence, the correct option is (C)
Question 7
1 / -0

An amount of 16 moles $H_2$ and 4 moles of $N_2$ is confined in a vessel of volume one liter. The vessel is heated to a constant temperature unit the equilibrium, the pressure was found to be $9/10^{Th}$ of the initial pressure. The value of $K_c$ for the reaction: $N_2(g) + 3H_2(g) \leftrightharpoons 2NH_3(g)$ is:

A
8100

B
$6.07 \times 10^{-4}$

C
1647.75

D
$8.99 \times 10^{-5}$

Solution
Question 8
1 / -0

When $20 ml$ of $0.2 M - DCl$ solution is mixed with $80 ml$ of $0.1 M - NaOD$ solution, $pD$ of the resulting solution becomes $13.6$ . The ionic product of heavy water, $D_{2}O$ , is

A
$10^{-15}$

B
$10^{-16}$

C
$4 \times 10^{-15}$

D
$4 \times 10^{-16}$
Question 9
1 / -0

A $40.0 ml$ solution of weak base, $BOH$ is titrated with $0.1 N - HCl$ solution. The $pH$ of the solution is found to be $10.0$ and $9.0$ after adding $5.0 ml$ and $20.0 ml$ of the acid, respectively. The dissociation constant of the base is ( $log 2 = 0.3$ )

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

B
$1 \times 10^{-5}$

C
$4 \times 10^{-5}$

D
$5 \times 10^{-5}$

Solution
Question 10
1 / -0

The process: $2A(g) \rightleftharpoons A_2(g)$ has $K_p = 8 \times 10^8 atm^{-1} .$ If 'A' atoms are taken at 1 atm pressure, what should be the equilibrium pressure of 'A' ?

A
0

B
$2.5 \times 10^{-5} atm$

C
$4 \times 10^4 atm$

D
$$ 1.25 \times 10
^{-5} $$

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Equilibrium Test - 71

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Equilibrium Test - 71

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Question 1

40.540.540.5 atm

818181 atm

33.7533.7533.75 atm

6.76.7 6.7 atm

Solution

Question 2

19 \dfrac {1}{9} 91​

181 \dfrac {1}{81} 811​

127 \dfrac {1}{27} 271​

13 \dfrac {1}{3} 31​

Solution

Question 3

no change

the new molar ratio becomes 1:1:51 : 1 :51:1:5

the new molar ratio becomes 2:2:52 : 2 : 52:2:5

the new molar ratio becomes 2:2:52 2 : 2 : 5\sqrt 2 2:2:52​

Solution

Question 4

concentration of SO2 SO_2 SO2​ is increased

more Cl2 Cl_2 Cl2​ is formed

concentration of all change

concentration will not change

Solution

Question 5

more ice will be formed

ice will sublime

more water will be formed

equilibrium will not disturb

Solution

Question 6

A(s)+2B(g)⇋C(g)+D(g);ΔH=−ve A(s) +2B(g) \leftrightharpoons C(g) +D(g); \Delta H = -ve A(s)+2B(g)⇋C(g)+D(g);ΔH=−ve

A(g)+2B(g)⇋2C(g)+D(g);ΔH=+ve A(g) +2B(g) \leftrightharpoons 2C(g) +D(g); \Delta H = +ve A(g)+2B(g)⇋2C(g)+D(g);ΔH=+ve

2A(g)+B(g)⇋2C(g)+D(s);ΔH=+ve 2A(g) +B(g) \leftrightharpoons 2C(g) +D(s); \Delta H = +ve 2A(g)+B(g)⇋2C(g)+D(s);ΔH=+ve

2A(s)+2B(g)⇋2C(g)+2D(g);ΔH=−ve 2A(s) +2B(g) \leftrightharpoons 2C(g) +2D(g); \Delta H = -ve 2A(s)+2B(g)⇋2C(g)+2D(g);ΔH=−ve

Solution

Question 7

8100

6.07×10−4 6.07 \times 10^{-4} 6.07×10−4

1647.75

8.99×10−5 8.99 \times 10^{-5} 8.99×10−5

Solution

Question 8

10−1510^{-15}10−15

10−1610^{-16}10−16

4×10−154 \times 10^{-15}4×10−15

4×10−164 \times 10^{-16}4×10−16

Question 9

2×10−52 \times 10^{-5}2×10−5

1×10−51 \times 10^{-5}1×10−5

4×10−54 \times 10^{-5}4×10−5

5×10−55 \times 10^{-5}5×10−5

Solution

Question 10

0

2.5×10−5atm2.5 \times 10^{-5} atm 2.5×10−5atm

4×104atm 4 \times 10^4 atm 4×104atm

$$ 1.25 \times 10^{-5} $$

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$40.5$ atm

$81$ atm

$33.75$ atm

$6.7$ atm

$\dfrac {1}{9}$

$\dfrac {1}{81}$

$\dfrac {1}{27}$

$\dfrac {1}{3}$

the new molar ratio becomes $1 : 1 :5$

the new molar ratio becomes $2 : 2 : 5$

the new molar ratio becomes $2 : 2 : 5\sqrt 2$

concentration of $SO_2$ is increased

more $Cl_2$ is formed

$A(s) +2B(g) \leftrightharpoons C(g) +D(g); \Delta H = -ve$

$A(g) +2B(g) \leftrightharpoons 2C(g) +D(g); \Delta H = +ve$

$2A(g) +B(g) \leftrightharpoons 2C(g) +D(s); \Delta H = +ve$

$2A(s) +2B(g) \leftrightharpoons 2C(g) +2D(g); \Delta H = -ve$

$6.07 \times 10^{-4}$

$8.99 \times 10^{-5}$

$10^{-15}$

$10^{-16}$

$4 \times 10^{-15}$

$4 \times 10^{-16}$

$2 \times 10^{-5}$

$1 \times 10^{-5}$

$4 \times 10^{-5}$

$5 \times 10^{-5}$

$2.5 \times 10^{-5} atm$

$4 \times 10^4 atm$

$$ 1.25 \times 10
^{-5} $$