Equilibrium Test

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Equilibrium Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

The number of hydronium ions in $$1 ml$$ of an aqueous solution of $$pH$$ $$12.0$$ at $$25^{\circ}C$$ is

A
$$0.01$$

B
$$10^{-12}$$

C
$$6.02 \times 10^{8}$$

D
$$6.02 \times 10^{11}$$

Solution
Question 2
1 / -0

$$pH$$ of $$0.01 M - (NH_{4})_{2}SO_{4})$$ and $$0.02 M - NH_{4}OH$$ buffer ($$pK_{a}$$ of $$NH_{4}^{+} = 9.26$$) is

A
$$9.26 + log 2$$

B
$$9.26 - log 2$$

C
$$4.74 +log 2$$

D
$$9.26$$

Solution
Question 3
1 / -0

For the reaction : $$ PCl_5(g) \leftrightharpoons PCl_3(g) + Cl_2(g) $$, if initial moles of $$ PCl_5 $$ is 'x', a is the degree of dissociation and P is total pressure at equilibrium , then $$ P_{PCl_3 } . P^{-1} $$ is equal to

A
$$ \dfrac { x}{1 + x} $$

B
$$ \dfrac { ax}{a+ x} $$

C
$$ \dfrac { a}{1 + a} $$

D
$$ \dfrac { a}{x+ ax} $$

Solution
Question 4
1 / -0

$$ K_p $$ for the reaction : $$ N_2O_4(g) \leftrightharpoons NO_2 $$ is 640 mm at 775 K. The percentage dissociation of $$ N_2O_4$$ at equilibrium pressure of 160 m is:

A
$$ \dfrac {100}{ \sqrt 2 } $$

B
$$50$$

C
$$ \dfrac {50}{ \sqrt 2 } $$

D
$$ 10 \sqrt 2 $$

Solution
Question 5
1 / -0

$$pH$$ of water is $$7.0$$ at $$25^{\circ}C$$. If water is heated to $$70^{\circ}C$$, the

A
$$pH$$ will decrease and the sample becomes acidic.

B
$$pH$$ will increase but the sample will remain neutral.

C
$$pH$$ will remain constant as $$7$$.

D
$$pH$$ will decrease but the sample will remain neutral.

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 $$25^{\circ}C$$

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, $$[H^+]>10^{-7}M$$ and $$pH$$ is less than 7 but the solution will remain neutral because nothing is added to water to change the concentration of $$H^+$$ or $$OH^-$$ ion.

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

Nitrogen dioxide, $$NO_{2}$$, exists in equilibrium with dinitrogen tetroxide, $$N_{2}O_{4}$$.
$$2NO_{2}(g)\rightleftharpoons N_{2}O_{4}(g)\ \triangle H = -57kJ\ mol^{-1}$$
Which conditions give the greatest percentage of $$N_{2}O_{4}(g)$$ at equilibrium?

A
Pressure - high, temperature - high

B
Pressure - high, temperature - low

C
Pressure - low, temperature - high

D
Pressure - low, temperature - low

Solution

Since the reaction is exothermic, the high pressure and low temperature will favour the reaction.
Question 7
1 / -0

For the reaction : $$ XY_2 (g) \leftrightharpoons XY(g) + Y (g) $$ the reaction is started with initial pressure of $$ XY_2 $$, $$600$$ mmHg. The total pressure for gases at equilibrium is 800 mmHg. Assuming volume and temperature of the system remains constant, the value of the $$ K_p $$ is

A
$$100$$ atm

B
$$100$$ mm Hg

C
$$0.01$$ atm

D
$$400$$ mm Hg

Solution
Question 8
1 / -0

The reaction between sulfur dioxide and oxygen is reversible.
$$2SO_2+O_2\rightleftharpoons 2SO_3$$
$$\Delta H^{\bigoplus}=-196$$ kJ $$mol^{-1}$$
Which conditions of pressure and temperature favour the reverse reaction?

Pressure Temperature
A high high
B high low
C low high
D low low

A
A

B
B

C
C

D
D

Solution

$$\text{The above reaction is a endothermic reaction since Change in enthalpy is negative.}$$
$$\text{1.High temperature is favourable for endothermic reaction.}$$
$$\text{2.Number of gas mole in product side is less than reactant so low pressure is favourable.}$$

$$\text{option C is correct.}$$
Question 9
1 / -0

Saccharin ($$K_{a} = 2 \times 10^{-12}$$) is a weak acid represented by formula, $$HSac$$. A $$4 \times 10^{-4}$$ mole amount of saccharin is dissolved in $$200 ml$$ water of $$pH$$, $$3.0$$. Assuming no change in volume, the concentration of $$Sac^{-1}$$ ions in the resulting solution at equilibrium is

A
$$4 \times 10^{-12} M$$

B
$$2 \times 10^{-12} M$$

C
$$8 \times 10^{-13} M$$

D
$$6.32 \times 10^{-8} M$$

Solution
Question 10
1 / -0

For the equilibrium $$PCl_5 \rightleftharpoons PCl_3 + Cl_2$$, $$K_p =\dfrac{\alpha^2}{(1 - \alpha) V}$$, temperature remaining constant,

A
$$K_p$$ will increase with the increase in volume

B
$$K_p$$ will increase with the decrease in volume

C
$$K_p$$ will not change with the change in volume

D
$$K_p$$ may increase or decrease with the change in volume

Solution

$$PCl_3 \rightleftharpoons \ PCl_3 + Cl_2, \ \ Kp = \cfrac{\alpha^2}{(1 - \alpha) V}$$

As volume changes, $$\alpha$$ will affect the equilibrium position.

Hence, $$K_p$$ will not change.