Equilibrium Test

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Equilibrium Test - 19

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Question 1
1 / -0

Assertion: On addition of $$NH_4Cl$$ to $$NH_4OH, pH$$ decreases but remains greater than 7.
Reason: Addition of $$\overset{\circleddash}{N}H_4$$ ion decreases ionisation of $$NH_4$$, thus [OH] decreases and also pH decreases.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

C
Assertion is correct but Reason is incorrect

D
Assertion is incorrect but Reason is correct

E
Both Assertion and Reason are incorrect

Solution

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

So addition of $$CH_3COONa$$ to $$CH_3COOH$$ increases the pH of solution but addition of $$NH_4Cl$$ to $$NH_4OH$$ decreases the pH of solution.
Question 2
1 / -0

Assertion: On cooling in a freezing mixture, colour of the mixture turns to pink from deep blue for a reaction.
$$CO(H_2O)_6\,^{2+}(aq)+4Cl^{\circleddash}(aq)\rightleftharpoons CoCl_4\,^{2-}(aq)+6H_2O(l)$$
(Pink) (Blue)
Reason: The reaction is endothermic in forward reaction, so on cooling the reaction, deep blue colour appears.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

C
Assertion is correct but Reason is incorrect

D
Assertion is incorrect but Reason is correct

E
Both Assertion and Reason are incorrect

Solution

The reaction $$ [Co(H_2O)_6]^{2+}(aq) + 4Cl^–(aq) \rightleftharpoons [CoCl_4]^{2–}(aq) + 6H_2O(l)$$ is endothermic. The shift of octahedral to tetrahedral coordination cause the colour change from pink to blue color that occurs when the second chloride ion enters the coordination sphere. Therefore, in accordance with Le Chatelier’s principle, when the temperature is raised, the position of the equilibrium will move to the right, forming more of the blue complex ion at the expense of the pink species. And when mixture freezes it supports backward reaction forming more pink complex.
Question 3
1 / -0

When $$NaN{O}_{3}$$ is heated in a closed vessel, oxygen is liberated and $$NaN{O}_{2}$$ is left behind. At equilibrium:

A
addition of $$NaN{O}_{2}$$ favours reverse reaction

B
addition of $$NaN{O}_{2}$$ favours forward reaction

C
increasing temperature favours forward reaction

D
decreasing pressure favours reverse reaction

Solution

$$NaN{O}_{3}\left(s\right) \rightleftharpoons NaN{O}_{2}\left(s\right) + \displaystyle\frac{1}{2}{O}_{2}\left(g\right); \Delta H = +ve$$

Also, $${K}_{p} = {\left({P}_{{O}_{2}}\right)}^{{1}/{2}}$$ and thus, addition of $$NaN{O}_{3}$$ or $$NaN{O}_{2}$$ has no effect, but increasing temperature favours forward reaction.
Question 4
1 / -0

Assertion: $$pH$$ value of $$HCN$$ solution decreases when $$NaCN$$ is added to it
Reason: $$NaCN$$ provides a common ion $$CN^{\circleddash}$$ to $$HCN$$

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

C
Assertion is correct but Reason is incorrect

D
Assertion is incorrect but Reason is correct

E
Both Assertion and Reason are incorrect

Solution

$$(A)$$ is wrong because the addition of $$NaCN$$ to $$HCN$$, due to common ion $$(CN^{\circleddash})$$, the degree of dissociation of $$HCN$$ is suppressed and hence less $$[H^{\oplus}]$$ and increase in $$pH.$$
Question 5
1 / -0

The equilibrium constant K for the reaction:
$$2HI(g)\rightleftharpoons H_2(g)+I_1(g)$$ at room temp is 2.85 and that at 698 K is $$1.4\times 10^{-2}$$. This implies that the forward reaction is:

A
exothermic

B
endothermic

C
exergonic

D
unpredictable

Solution

With the increase of temperature, k value decreases, so that forward reaction decreases with increase of temperature. This implies that reaction will proceed in forward direction with decrease of temperature, i.e., heat is liberated and hence forward reaction is exothermic.
Question 6
1 / -0

The equilibrium constant for a reaction,
$$A+B\rightleftharpoons C+D$$ is $$1.0\times 10^{-2}$$ at $$298$$ and is $$2.0$$ at $$373$$ K. The chemical process resulting in the formation of C and D is :

A
Exothemic

B
Endothermic

C
unpredictable

D
none

Solution

Since K increases on increasing temperature, so the reaction will go forward by increasing temperature and hence is endothermic.
Question 7
1 / -0

The system $$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$$ attains equilibrium. If the equilibrium concentration of $$PCl_3(g)$$ is doubled, the concentration of $$Cl_2(g)$$ would become:

A
$$\dfrac{1}{4}$$ its original value

B
$$\dfrac{1}{2}$$ its original value

C
twice its original value

D
unpredictable

Solution

$$PCl_5\rightleftharpoons PCl_3+Cl_2$$
If the concentration of $$PCl_3$$ is doubled, it only predicts, that reaction will proceed backward. (Le-Chatlier's principle). So the concentration of
$$Cl_2$$, is unpredictable.
Question 8
1 / -0

In the gaseous equilibrium $$A+2B\rightleftharpoons C+Heat$$, the forward reaction is favoured :

A
Low P, High T

B
Low P, Low T

C
High P, Low T

D
High P, High T

Solution

$$A+2B\rightleftharpoons C+Heat$$
The equation shows, that it is exothermic reaction. Since the heat is released in the reaction, so the reaction is favoured in forward direction at low temperature.
$$\Delta n=1-(2+1)=-2$$
Since the number of moles decreases, $$\therefore$$ the forward reaction is favoured at high pressure.
Question 9
1 / -0

For a hypothetical reaction of the kind
$$AB_2(g)+\frac {1}{2}B_2(g)\rightleftharpoons AB_3(g);\Delta H=-x kJ$$ More $$AB_3$$ could be produced at equilibrium by :

A
using a catalyst

B
removing some of $$B_2$$

C
increasing the temperature

D
increasing the pressure

Solution

Since the number of moles decreases. So reaction is favoured forward by increasing pressure,
$$\Delta n=1-\left (1+\frac {1}{2}\right )=-\frac {1}{2}$$
Question 10
1 / -0

The solubility of $$CO_2$$ in water increases with :

A
increase in temperature

B
reduction of gas pressure

C
increase in gas pressure

D
increase in volume

Solution

When carbon dioxide reacts with water it forms carbonic acid. Increasing the pressure of carbon dioxide makes the reaction feasible in the forward direction and hence solubility of $$ CO_2$$ increases.

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Equilibrium Test - 19

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Equilibrium Test - 19

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Question 1

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

Both Assertion and Reason are incorrect

Solution

Question 2

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

Both Assertion and Reason are incorrect

Solution

Question 3

addition of $$NaN{O}_{2}$$ favours reverse reaction

addition of $$NaN{O}_{2}$$ favours forward reaction

increasing temperature favours forward reaction

decreasing pressure favours reverse reaction

Solution

Question 4

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

Both Assertion and Reason are incorrect

Solution

Question 5

exothermic

endothermic

exergonic

unpredictable

Solution

Question 6

Exothemic

Endothermic

unpredictable

none

Solution

Question 7

$$\dfrac{1}{4}$$ its original value

$$\dfrac{1}{2}$$ its original value

twice its original value

unpredictable

Solution

Question 8

Low P, High T

Low P, Low T

High P, Low T

High P, High T

Solution

Question 9

using a catalyst

removing some of $$B_2$$

increasing the temperature

increasing the pressure

Solution

Question 10

increase in temperature

reduction of gas pressure

increase in gas pressure

increase in volume

Solution

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