Equilibrium Test

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Equilibrium Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

The concentration of $$\left[ {{H^ + }} \right]$$ and concentration of $$\left[ {{OH^ + }} \right]$$ of a $$0.1\,M$$ aqueous solution of $$2\% $$ ionised weak acid is [ionic product of water $$ = 1 \times {10^{ - 14}}$$]

A
$$2 \times {10^{ - 3}}M$$ and $$5 \times {10^{ - 12}}M$$

B
$$1 \times {10^{ - 3}}M$$ and $$3 \times {10^{ - 11}}M$$

C
$$0.02 \times {10^{ - 3}}M$$ and $$5 \times {10^{ - 11}}M$$

D
$$3 \times {10^{ - 2}}M$$ and $$4 \times {10^{ - 13}}M$$

Solution

$$HA $$ $$ \rightleftharpoons $$ $$H^{+} + A^{-} $$
t=0: $$c$$
t=t: $$c(1 - \alpha) $$ $$c \alpha $$ $$c \alpha$$
$$ \therefore $$ $$[H^{+}] = c \alpha = 0.1 \times 0.02 = 2 \times 10^{-3} $$

Hence $$[OH^{-}] = \dfrac{10^{-14}}{2 \times 10^{-3}} = 5 \times 10^{-12} $$

Hence, Option "B" is the correct answer.
Question 2
1 / -0

The reaction $${ 2HI }_{ \left( g \right) }\rightleftharpoons { H }_{ 2\left( g \right) }+{ I }_{ 2\left( g \right) }$$ is initiated within a flask with 0.2 atm pressure of HI. If the partial pressure of $${ HI }_{ \left( g \right) }$$ at equilibrium is 0.04 atm the equilibrium constant of the reaction is

A
4.0

B
16.0

C
25.0

D
20.0

Solution

at equilibrium ,
pressure of HI = 0.2 - 2x = 0.04
2x = 0.2 - 0.04 = 0.16
x = 0.08 atm

so, pressure of H2 and I2 at equilibrium is 0.08 atm .
e.g., at equilibrium ,

Kp = 0.08atm × 0.08atm/(0.04atm)²= 4

hence, Kp = 4
Question 3
1 / -0

Addition of sodium acetate solution to acetic acid causes the following change-

A
pH increases

B
pH decreases

C
pH remains unchanged

D
pH becomes 7

Solution

Dissociation of $$ CH_{3}COOH $$ is suppressed by the addition of sodium acetate $$ ( CH_{3}COONa) $$ due to common ion $$ (CH_{3}COO^{-}) $$ effect. The $$ [H^+]$$ decreases raising the pH of the acid solution.

Hence, pH will increase.

Therefore, option A is correct.
Question 4
1 / -0

Predict the direction in which the following equilibrium will shift backward.
$$\underset{(g)}{N_2} + \underset{(g)}{3H_2} \Leftrightarrow \underset{(g)}{2NH_3 \,\, \Delta H} = -100 \, Joule$$

A
increasing Temperature

B
increasing pressure

C
addition of catalyst

D
on addition of $$NH_{3}$$

Solution

as the given reaction has enthalpy of formation negative,thus reaction is exothermic.
Thus on increasing temperature equilibrium will shift in the backward direction as reaction is exothermic.
on increasing pressure equilibrium will shift in backward direction.
Addition of catalyst does not effect equilibrium positions.
Addition of NH3 will lead in the shifting of equilibrium in backward direction.
Question 5
1 / -0

Which of the following pair will show common ion effect?

A
HCN +KOH

B
HBr +AgBr

C
HCOOH + $$ CH_3COOH $$

D
KOH +AgOH
Question 6
1 / -0

In which manner will increases of pressure affect the following equation ?
$$C(s)+{ H }_{ 2 }O(g)\rightleftharpoons CO(g)+{ H }_{ 2 }(g)$$

A
shift in the forward direction

B
shift in the reverse direction

C
increase in the yield of hydrogen

D
no effect

Solution

The forward reaction is accompanied by the increase in the number of gaseous moles. Hence increase of pressure will favour the reverse reaction.

A is corret option
Question 7
1 / -0

Which changes increase the rate of reaction of an exothermic reaction?
1.Temperature increase
2.Temperature decrease

A
only 1

B
1 and 2 both

C
only 2

D
none

Solution

$$\text{In a exothermic reaction heat relesed from the reaction}$$
$$\text{So decreasing the temperature of the reaction increases the rate of the reaction}$$
Question 8
1 / -0

For a reaction: A$$\rightleftharpoons$$B; $$\Delta H^o < 0$$ and $$\Delta S^o > 0$$. Which of the following plots describes accurately the variation of logarithmic and equilibrium constant with temperature?

A

B

C

D
Question 9
1 / -0

Solubility of $$MX_{ 2 }$$ type electrolytes is $$0.5\times 10^{ -4 } mol/L$$, Then find out $${ K }_{ sp }$$ of electrolytes.

A
$$5\times 10^{ -12 }$$

B
$$25\times 10^{ -10 }$$

C
$$1\times 10^{ -13 }$$

D
$$5\times 10^{ -13 }$$

Solution

For $$MX_{2}$$ type of electrolyte
$$MX_{2}\rightleftharpoons M^{+2} + 2X^{-}$$. So if the solubility is $$S$$ then the solubility product would be $$4S^{3}$$
Here, solubility S= $$0.5 \times 10^{-4}mol/L$$
So, solubility product $$K_{sp}=4\times (0.5 \times10^{-4})^{3}$$
=$$5 \times 10^{-13}$$ So the correct option is D
Question 10
1 / -0

In a $$5.76$$ L vessel, $$0.5$$ moles of $$ H_2 $$ gas and $$0.5$$ moles of $$ I_2 $$ vapours are allowed to react to from HI(g) at $$ 447^0 C $$, then total pressure of gases at equilibrium would be ( R =0.08 L-atm /K-mol )

A
$$20$$ atm

B
$$10$$ atm

C
$$5$$ atm

D
$$1$$ atm

Solution