Equilibrium Test

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Equilibrium Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

Given exothermic reaction,

$$[CoCl_4]^{2-} (aq)+6H_2O \rightleftharpoons [Co(H_2O)_6]^{2+} + 4Cl^- (aq)$$

Which one of the following will decrease the equilibrium concentration of $$[CoCl_4]^{2-}$$?

A
Addition of $$HCl$$

B
Addition of $$ [Co(H_2O)_6]^{2+}$$

C
The solution is diluted with water

D
The temperature is increased
Question 2
1 / -0

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

A
two times the original value

B
one half of its original value

C
$$2 \sqrt{2} $$ times to the original value

D
$$ \cfrac{1}{2 \sqrt{2}} $$ times the original value
Question 3
1 / -0

When ammonia is added to pure water, the concentration of which of the following species already present, will decrease?

A
$$OH^-$$

B
$$H_3O^+$$

C
$$NH_4^+$$

D
None of these
Question 4
1 / -0

For which salt the $$\mathrm { pH }$$ of its solution does not change with dilution?

A
$$\mathrm { NH } _ { 4 } \mathrm { Cl }$$

B
$$\mathrm { CH } _ { 3 } \mathrm { COONH } _ { 4 }$$

C
$$\mathrm { CH } _ { 3 } \mathrm { COONa }$$

D
None of these

Solution
Question 5
1 / -0

What is $$ [H^+] $$ in mol/L of a solution that is 0.20 M $$ CH_3CCONa $$ and 0.10 M in $$ CH_3COOH $$?
$$( K_a\ for\ CH_3COOH = 1.8 \times 10^{-5} ) $$

A
$$ 3.5 \times 10^{-4} $$

B
$$ 1.1 \times 10^{-5} $$

C
$$ 1.8 \times 10^{-5} $$

D
$$ 9.0 \times 10^{-6} $$

Solution
Question 6
1 / -0

The equilibrium of the reaction $$ N_2(g) + 3H_2(g) \leftrightharpoons 2NH_3(g) $$ will shift to the product side when

A
$$ K_p > 1 $$

B
$$ Q > K_p $$

C
$$ Q = K_p $$

D
$$ Q<K_p $$
Question 7
1 / -0

The $$pH$$ of an aqueous solution of sodium chloride at $$60^{\circ}C$$ is

A
$$7.0$$

B
$$> 7.0$$

C
$$<7.0$$

D
$$0$$

Solution

We know that at $$25^{\circ}C$$ dissociation constant $$K_w$$ of water is,

$$H_2O\rightleftharpoons H^++OH^-$$

$$[H^+][OH^-]=10^{-14}=K_w$$

So, $$[H^+]=[OH^-]=10^{-7}M$$

Thus $$pH=7$$ for $$H_2O$$ at $$25^{\circ}C$$

Adding $$NaCl$$ does not affect the $$pH$$ whatever be the temperature because sodium chloride is salt of strong acid and strong base. So, it is $$H^+$$ ion of water which is affecting the $$pH$$.

Dissociation of water is an endothermic process. With an increase in temperature dissociation constant of water increases. So the dissociation constant is more i.e., $$K_w>10^{-14}$$.

Therefore, $$[H^+]>10^{-7}$$ and $$pH$$ is less than 7.

Hence the correct option is (C).
Question 8
1 / -0

The dissociation constant of formic acid is $$0.00024$$. The hydrogen ion concentration in $$0.002 M \ HCOOH$$ solution is nearly

A
$$6.93 \times 10^{-4}M$$

B
$$4.8 \times 10^{-7}M$$

C
$$5.8 \times 10^{-4}M$$

D
$$1.4 \times 10^{-4}M$$
Question 9
1 / -0

The degree of dissociation of pure water at $$25^{\circ}C$$ is found to be $$1.8 \times 10^{-9}$$. The dissociation constant, $$K_{d}$$ of water, at $$25^{\circ}C$$ is

A
$$10^{-14}$$

B
$$1.8 \times 10^{-16}$$

C
$$5.56 \times 10^{-13}$$

D
$$1.8 \times 10^{-14}$$

Solution

Given:

Degree of dissociation of pure water, $$\alpha =1.8\times 10^{-9}$$

We know that,
Density of pure water = $$1\space g\space cm^{-3}=1000\space g\space L^{-1}$$
Molar mass of water = $$18\space g\space mol^{-1}$$

$$Concentration=\frac{number\space of\space moles}{Volume}=\frac{1000\space g\space L^{-1}}{18\space g\space mol^{-1}}=55.5\space M$$

Dissociation of water, $$H_2O\rightleftharpoons H^++OH^-$$

Dissociation constant, $$K_d=\frac{[H^+][OH^-]}{[H_2O]}=\frac{C\alpha\times C\alpha}{C-C\alpha}$$

As $$\alpha<<<1$$, $$[H_2O]$$ is not affected.
Therefore,

$$K_d=c\alpha^2=55.5\times (1.8\times 10^{-9})^2=1.79\times 10^{-16}\approx 1.8\times 10^{-16}$$

Hence, the correct option is (B).
Question 10
1 / -0

For a reversible reaction : $$ A +B \leftrightharpoons C $$ if the concentration of the reactions are doubled at a definite temperature, then equilibrium constant will

A
be doubled

B
be havled

C
be one fourth

D
remain same

Solution

For a reversible reaction :$$ A+B⇋C$$

Doubling the initial concentrations of reactants in equilibrium does not result in a change in $$K_c$$ value. $$K_c$$ remains the same at a particular temperature.

Therefore option $$D$$ is a correct option.

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Re-Attempt Weekly Quiz Competition

Equilibrium Test - 69

Result

Equilibrium Test - 69

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SHARING IS CARING

Question 1

Addition of $$HCl$$

Addition of $$ [Co(H_2O)_6]^{2+}$$

The solution is diluted with water

The temperature is increased

Question 2

two times the original value

one half of its original value

$$2 \sqrt{2} $$ times to the original value

$$ \cfrac{1}{2 \sqrt{2}} $$ times the original value

Question 3

$$OH^-$$

$$H_3O^+$$

$$NH_4^+$$

None of these

Question 4

$$\mathrm { NH } _ { 4 } \mathrm { Cl }$$

$$\mathrm { CH } _ { 3 } \mathrm { COONH } _ { 4 }$$

$$\mathrm { CH } _ { 3 } \mathrm { COONa }$$

None of these

Solution

Question 5

$$ 3.5 \times 10^{-4} $$

$$ 1.1 \times 10^{-5} $$

$$ 1.8 \times 10^{-5} $$

$$ 9.0 \times 10^{-6} $$

Solution

Question 6

$$ K_p > 1 $$

$$ Q > K_p $$

$$ Q = K_p $$

$$ Q<K_p $$

Question 7

$$7.0$$

$$> 7.0$$

$$<7.0$$

$$0$$

Solution

Question 8

$$6.93 \times 10^{-4}M$$

$$4.8 \times 10^{-7}M$$

$$5.8 \times 10^{-4}M$$

$$1.4 \times 10^{-4}M$$

Question 9

$$10^{-14}$$

$$1.8 \times 10^{-16}$$

$$5.56 \times 10^{-13}$$

$$1.8 \times 10^{-14}$$

Solution

Question 10

be doubled

be havled

be one fourth

remain same

Solution

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