Equilibrium Test

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Equilibrium Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

It accepts a proton. It is called as:

A
a Bronsted acid

B
a Bronsted base

C
a strong acid

D
a weak base

Solution

A Bronsted acid can accept a proton $$(H^+)$$. For example a basic salt, such as $$Na^{\oplus}H^ {\circleddash}$$, can take up a $$H^+$$ from $$H_2O$$ to form $$OH^{\circleddash}$$ ions.
$$F^ {\circleddash}(aq)+H_2O(l) \rightleftharpoons HF (aq)+OH^ {\oplus}(aq)$$
Question 2
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What is a buffer?

A
A solution that contains equal amounts of a weak acid and its conjugate base, or a weak base and its conjugate acid.

B
A solution that contains both a strong acid and a base.

C
A solution that contains an acid.

D
A solution that contains equal amounts of a strong acid and its conjugate base, or a strong base and its conjugate acid.

E
A solution that contains a base.

Solution

A buffer is a solution that contains equal amounts of a weak acid and its conjugate base, or a weak base and its conjugate acid.
Such solution resists the change in pH when small amount of an acid or a base is added.
For example, a mixture of acetic acid and sodium acetate acts as an acid buffer solution. Similarly, a mixture of ammonium hydroxide and ammonium chloride is a basic buffer solution.
Question 3
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$${N}_{2}(g)+3{H}_{2}(g)\rightleftharpoons 2{NH}_{3}(g)+heat$$
Given the reaction shown above, which change will not shift the equilibrium position for the reaction?

A
Addition of $${N}_{2}$$

B
Addition of $${NH}_{3}$$

C
Cooling the reaction system

D
Addition of an iron catalyst

E
Decreasing the pressure

Solution

The given reaction is :-
$${ N }_{ 2 }\left( g \right) +3{ H }_{ 2 }\left( g \right) \rightleftharpoons 2{ NH }_{ 3 }\left( g \right) +heat$$
Addition of Iron catalyst will not shift the equilibrium position of the reaction. The catalyst only help the reaction to reach equilibrium quickly by altering the reaction pathway by lowering the activation energy. It increases the rate of both forward & backward reaction in same amounts so as to attain equilibrium quickly.
Question 4
1 / -0

What is the $${K}_{eq}$$ for the following reaction?
$${H}_{2}(g)+{I}_{2}(g)\rightleftharpoons 2HI(g)$$

A
$$\cfrac { \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] }{ 2\left[ HI \right] } $$

B
$$\cfrac { { \left[ HI \right] }^{ 2 } }{ \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] } $$

C
$$\cfrac { \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] }{ { \left[ HI \right] }^{ 2 } } $$

D
$$\cfrac { 2\left[ HI \right] }{ \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] } $$

E
$$\cfrac { \left[ { H }_{ 2 } \right] +\left[ { I }_{ 2 } \right] }{ { \left[ HI \right] }^{ 2 } } $$

Solution

The given reaction is :-
$${ H }_{ 2 }\left( g \right) +{ I }_{ 2 }\left( g \right) \rightleftharpoons 2HI\left( g \right) $$
$$\therefore \quad { K }_{ C }=\dfrac { { \left[ HI \right] }^{ 2 } }{ \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] } $$
Question 5
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Statement I: Water makes a good buffer.
Statement II: A good buffer will resist changes in pH.

A
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1

B
Both the Statement 1 and Statement 2 are correct and Statement 2 is NOT the correct explanation of Statement 1

C
Statement 1 is correct but Statement 2 is not correct

D
Statement 1 is not correct but Statement 2 is correct

E
Both the Statement 1 and Statement 2 are not correct.

Solution

Option D is the correct answer.
Water is not a good buffer, because its pH is very sensitive to addition of any acid or base.
A good buffer will resist a change in pH upon addition of an acid or a base.
Thus, statement 1 is not correct but statement 2 is correct.
Question 6
1 / -0

Addition of water to this solution will not change $$[H_3 O^+]$$.

A
Chemical pH indicator

B
Acid/base buffer

C
Anhydrous solution

D
Hypotonic solution

Solution

Addition of water to $$X$$ solution does not change the $$pH$$ of the solution, which means the concentration of the species present changes in such a way that the $$pH$$ remains the same. This is possible for acid/base buffer.
$$pH=pk_a+\log \cfrac{[salt]}{[acid]}$$
Since the volume changes the same for both salt and acid, the ratio $$\cfrac{[salt]}{[acid]}$$ remains the same and hence $$pH$$ also remains the same. Same in the case of basic buffer solution.
Question 7
1 / -0

For each pair of chemicals below, the amounts shown are dissolved, together, in water to make 1.0 L of solution. Which mixture will form a buffer solution?

A
0.2 mol of $$\displaystyle HCl$$ and 0.1 mol of $$\displaystyle { K }_{ 2 }{ SO }_{ 3 }$$

B
0.2 mol of $$\displaystyle NaOH$$ and 0.4 mol of $$\displaystyle HF$$

C
0.1 mol of $$\displaystyle HBr$$ and 0.1 mol of $$\displaystyle Ba{ \left( OH \right) }_{ 2 }$$

D
0.4 mol of $$\displaystyle HCl$$ and 0.2 mol of $$\displaystyle { NH }_{ 3 }$$

E
0.2 mol of $$\displaystyle NaOH$$ and 0.4 mol of $$\displaystyle HCl$$

Solution

A buffer solution is an aqueous solution consisting of a mixture of an acid and its conjugate base. Its $$pH$$ changes very little when a small amount of strong acid or base is added to it. Hence solution of $$NaOH /HF$$ is a buffer solution.
Question 8
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Directions For Questions

A chemist interested in the reactivity of iodine concentrates his study on the decomposition of gaseous hydrogen iodide. The reaction is $$2HI(g) \leftrightharpoons H_2(g) +I_2(g)$$.

...view full instructions

An increase in pressure in this reaction would:

A
produce more $$I^- (aq)$$

B
produce more $$H_2$$

C
not affect the system

D
drive it to the right

Solution

The given reaction is :-
$$2HI\left( g \right) \rightleftharpoons { H }_{ 2 }\left( g \right) +{ I }_{ 2 }\left( g \right) $$
This reaction takes place in either direction without change in no. of moles. So, a/c to Le Chatelier's principle, pressure will have no effect on this equilibrium.
Question 9
1 / -0

If the $${K}_{eq}$$ for a reversible reaction is slightly less than $$1$$, which of the flowwoing statements is true?

A
The formation of products is slightly favored.

B
The formation of products is greatly favored.

C
The formation of reactants is slightly favored.

D
The formation of reactants is greatly favored.

E
Neither the formation of products nor reactants is favored.

Solution

If equilibrium constant of reaction is less than $$1$$.
Then, $$\because \quad { K }_{ eq }=\dfrac { { K }_{ f } }{ { K }_{ b } } \quad ;\quad { K }_{ f }$$ and $${ K }_{ b }$$ are same as defined above.
Now, $${ K }_{ eq }<1\Rightarrow { K }_{ f }<{ K }_{ b }$$.
So, Rate of forward reaction is less than rate of backward reaction. So, less reactants will react to form products and equation will lie to the left.
Question 10
1 / -0

$$P{Cl}_{3}(g)+{Cl}_{2}(g)\rightleftharpoons P{Cl}_{5}(g)+energy$$
For the reaction above, what is the effect of increasing the pressure?

A
Increased production of products

B
Wild fluctuations in the amounts of reactants and products

C
Increased production of reactants

D
No impact on the equilibrium

Solution

The given reaction is:-

$${ PCl }_{ 3 }\left( g \right) +{ Cl }_{ 2 }\left( g \right) \rightarrow { PCl }_{ 5 }\left( g \right) +energy$$

In this reaction, the forward reaction is accompanied by a decrease of moles of gaseous species. So, if pressure on the system is increased then, as per Le Chatelier's principle, the equation shifts in direction in which a decrease in total no. of moles takes place i.e. in formation of products in this case.

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Equilibrium Test - 29

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Equilibrium Test - 29

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Question 1

a Bronsted acid

a Bronsted base

a strong acid

a weak base

Solution

Question 2

A solution that contains equal amounts of a weak acid and its conjugate base, or a weak base and its conjugate acid.

A solution that contains both a strong acid and a base.

A solution that contains an acid.

A solution that contains equal amounts of a strong acid and its conjugate base, or a strong base and its conjugate acid.

A solution that contains a base.

Solution

Question 3

Addition of $${N}_{2}$$

Addition of $${NH}_{3}$$

Cooling the reaction system

Addition of an iron catalyst

Decreasing the pressure

Solution

Question 4

$$\cfrac { \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] }{ 2\left[ HI \right] } $$

$$\cfrac { { \left[ HI \right] }^{ 2 } }{ \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] } $$

$$\cfrac { \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] }{ { \left[ HI \right] }^{ 2 } } $$

$$\cfrac { 2\left[ HI \right] }{ \left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] } $$

$$\cfrac { \left[ { H }_{ 2 } \right] +\left[ { I }_{ 2 } \right] }{ { \left[ HI \right] }^{ 2 } } $$

Solution

Question 5

Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1

Both the Statement 1 and Statement 2 are correct and Statement 2 is NOT the correct explanation of Statement 1

Statement 1 is correct but Statement 2 is not correct

Statement 1 is not correct but Statement 2 is correct

Both the Statement 1 and Statement 2 are not correct.

Solution

Question 6

Chemical pH indicator

Acid/base buffer

Anhydrous solution

Hypotonic solution

Solution

Question 7

0.2 mol of $$\displaystyle HCl$$ and 0.1 mol of $$\displaystyle { K }_{ 2 }{ SO }_{ 3 }$$

0.2 mol of $$\displaystyle NaOH$$ and 0.4 mol of $$\displaystyle HF$$

0.1 mol of $$\displaystyle HBr$$ and 0.1 mol of $$\displaystyle Ba{ \left( OH \right) }_{ 2 }$$

0.4 mol of $$\displaystyle HCl$$ and 0.2 mol of $$\displaystyle { NH }_{ 3 }$$

0.2 mol of $$\displaystyle NaOH$$ and 0.4 mol of $$\displaystyle HCl$$

Solution

Question 8

produce more $$I^- (aq)$$

produce more $$H_2$$

not affect the system

drive it to the right

Solution

Question 9

The formation of products is slightly favored.

The formation of products is greatly favored.

The formation of reactants is slightly favored.

The formation of reactants is greatly favored.

Neither the formation of products nor reactants is favored.

Solution

Question 10

Increased production of products

Wild fluctuations in the amounts of reactants and products

Increased production of reactants

No impact on the equilibrium

Solution

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