Equilibrium Test

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Equilibrium Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

The reaction, $$C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)$$ is at equilibrium in a closed vessel at $$1000\ K$$. The enthalpy change $$(\Delta H)$$ for the reaction is $$137.0$$ $$kJ.mol^{-1}$$. Which one of the following actions would shift the equilibrium to the right?

A
Decreasing the volume of the closed reaction vessel

B
Decreasing the temperature at which the reaction is performed

C
Adding an inert gas to the closed reaction vessel

D
Increasing the volume of the closed reaction vessel

Solution

On decreasing the volume of reaction, a reaction will move in backward direction due to lesser number of moles.

On decreasing the temperature of reaction, the reaction will move in backward direction as the reaction is endothermic reaction.

Adding an inert gas does not result in a shift as there is no change in concentration.

Increasing the volume of reaction will shift the equilibrium in forward direction due to the higher number of moles or due to decrease in concentration.
Question 2
1 / -0

Consider the reaction:
$${Cl}_{2}(aq)+{H}_{2}S(aq)\rightarrow S(s)+2{H}^{+}(aq)+2{Cl}^{-}(aq)$$

The rate equation for this reaction is
rate $$=k\left[ { Cl }_{ 2 } \right] \left[ { H }_{ 2 }S \right] $$

Which of these mechanisms is/are consistent with this rate equation?
A. $${Cl}_{2}+{H}_{2}S\rightarrow {H}^{+}+{Cl}^{-}+{Cl}^{+}+{HS}^{-}$$ (slow)
$${Cl}^{+}+{HS}^{-}\rightarrow H^++{Cl}^{-}+S$$ (fast)

B. $${H}_{2}S \rightleftharpoons {H}^{+}+{HS}^{-}$$ (fast equilibrium)
$${Cl}_{2}+{HS}^{-}\rightarrow 2{Cl}^{-}+{H}^{+}+S$$ (Slow)

A
$$B$$ only

B
Both $$A$$ and $$B$$

C
Neither $$A$$ nor $$B$$

D
$$A$$ only

Solution

The slow step is the rate-determining step.
From the given rate law expression (Rate$$=k\left[ { Cl }_{ 2 } \right] \left[ { H }_{ 2 }S \right] $$), it can be said that the slow step should involve $$1$$ molecule of $${Cl}_{2}$$ and $$1$$ molecule of $${H}_{2}S$$.

If we consider mechanism (A) we find that $$1$$ molecule of $${Cl}_{2}$$ and $$1$$ molecule of $${H}_{2}S$$ participate in the reaction. Hence, the rate law expression will be
Rate$$=k\left[ { Cl }_{ 2 } \right] \left[ { H }_{ 2 }S \right] $$

If we consider mechanism (B) we find
Rate$$=k\left[ { Cl }_{ 2 } \right] \left[ { HS }^{ - } \right] ...(i)$$

For equation
$${ H }_{ 2 }S\rightleftharpoons { H }^{ + }+{ HS }^{ - }$$

$$K=\cfrac { \left[ { H }^{ + } \right] \left[ { HS }^{ - } \right] }{[ { H }_{ 2 }S] } $$

or $$\left[ { HS }^{ - } \right] =\cfrac { K\left[ { H }_{ 2 }S \right] }{ [{ H }^{ + }] } $$

Substituting this value in equation (i) we find

Rate, $$r=k[Cl_2]\dfrac{K[H_2S]}{[H^+]}=k'\dfrac{[Cl_2][H_2S]}{[H^+]}$$

This is different from given rate law expression Rate$$=k\left[ { Cl }_{ 2 } \right] \left[ { H }_{ 2 }S \right] $$.

Hence only, mechanism ($$A$$) is consistent with the given rate equation.
Question 3
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An excess of $$Ag_2CrO_4(s)$$ is added to a $$5\times 10^{-3}M$$ $$K_2CrO_4$$ solution. The concentration of $$Ag^+$$ in the solution is closest to.[Solubility product for $$Ag_2CrO_4=1.1\times 10^{-12}$$]

A
$$2.2\times 10^{-10}$$M

B
$$1.5\times 10^{-5}$$M

C
$$1.0\times 10^{-6}$$M

D
$$5.0\times 10^{-3}$$M

Solution

Since, $$K_2CrO_4$$ is a strong electrolyte so the concentration of chromate ion will be almost equal to the concentration of $$K_2CrO_4$$.

$$Ag_2CrO_4\leftrightharpoons 2Ag^+ + CrO_4^{2-}$$
$$S$$ $$2S$$ $$5\times 10^{-3}\ M$$

Now,
$$K_{sp}=[Ag^+]^2[CrO_4^{2-}]$$

$$1.1\times 10^{-12}=[Ag^+]^2[5\times 10^{-3}]$$

$$[Ag^+]=1.5\times 10^{-5}$$
Question 4
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$$Au(s)\rightleftharpoons Au(l)$$

Above equilibrium is favoured at:

A
high pressure low temperature

B
high pressure high temperature

C
low pressure high temperature

D
low pressure low temperature

Solution

Using Le Chaterlier's principle:
when a system in equillibrium experiences a disturbance (such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state by opposing the disturbance.
For the process:
$$Au(s)\rightleftharpoons Au(l)$$
It is an endothermic process, therefore an increase in temperature will move the equilibrium forward.
Also, upon melting, volume of metals increases and the equilibrium can be shifted forward by reducing the melting point of metals. This can be achieved by reducing the pressure.
Therefore low pressure and high temperature favours the process.
Option C is correct.
Question 5
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A gas $$X$$ when dissolved in water heat is evolved. Then solubility of $$X$$ will increase:

A
low pressure, high temperature

B
low pressure, low temperature

C
high pressure, high temperature

D
high pressure, low temperature

Solution

$$X(g)+{ H }_{ 2 }O\rightleftharpoons X(aq);\quad -\Delta H$$
Using Le Chaterlier's principle:
when a system in equillibrium experiences a disturbance (such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state by opposing the disturbance.
An exothermic process goes into forward direct when temperature is decreased and since in the given process the gas goes into water thereby results in decrease of pressure, hence an increase in pressure will shift the equillibrium forward.
Therefore option D is correct.
Question 6
1 / -0

The solubility product constant $$Ksp$$ of $$Mg(OH)_{2}$$ is $$9.0\times 10^{-12}$$. If a solution is $$0.010\ M$$ with respect to $$Mg^{2+}$$ ion. What is the maximum hydroxide ion concentration which could be present without causing the precipitation of $$Mg(OH)_{2}$$?

A
$$1.5\times 10^{-7}M$$

B
$$3.0\times 10^{-7}M$$

C
$$1.5\times 10^{-5}M$$

D
$$3.0\times 10^{-5}M$$

Solution

$$Ksp(Mg(OH)_2)=9.0\times 10^{-12}$$
$$(Mg(OH)_2 \leftrightharpoons Mg^{2+}+2[OH]^-$$
$$Ksp=[Mg^{2+}][OH^-]^2$$
$$9\times 10^{-12}=(10^{-12})(OH^-)^2$$
$$[OH^-]^2=3^2\times (10^{-5})^2$$
$$[OH^-]=3.0\times 10^{-5}M$$
Maximum Hydroxide-ion concentration.
Question 7
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$$1$$ litre buffer solution $$\left( HA+NaA \right) pH=4$$ , is mixed with $$2$$ litre buffer solution $$\left( HA+NaA \right) pH=4.3$$. $$pH$$ of resulting solution if both solutions are initially $$0.1M$$ $$MA$$.
[Given $${ \left( { K }_{ a } \right) }_{ HA }={ 10 }^{ -4 };\log { 2 } =0.3;\log { 3 } =0.48;\log { 5 } =0.7\quad $$]

A
$$4.2$$

B
$$4.22$$

C
$$4.15$$

D
$$6.4$$

Solution
Question 8
1 / -0

Ammonia gas at $$15$$ atm is introduced in a rigid vessel at $$300$$K. At equilibrium, the total pressure of the vessel is found to be $$40.11$$ atm at $$300^0$$C. The degree of dissociation of $$NH_3$$ will be :

A
$$0.6$$

B
$$0.4$$

C
Unpredictable

D
None of these

Solution

$$ \begin{array}{l} P_{1}=15,9+m \\ T_{1}=300 \mathrm{~K} \\ P_{1}=? \\ T_{2}=300+273=573 \mathrm{~K} \end{array} $$
$$ \begin{array} \text { By } P V=n R T \\ \Rightarrow \quad \frac{P V}{n T}=R+\text { const. } \\ \Rightarrow \quad \frac{P_{1} V_{1}}{n_{1} T_{1}}=\frac{P_{2} V_{2}}{n_{2} T_{2}} \end{array} $$
$$ \begin{aligned} \Rightarrow \quad \frac{P_{1}}{T_{1}} &=\frac{P_{2}}{T_{2}} \\ \Rightarrow \quad \frac{15}{500} &=\frac{P_{2}}{573} \\ P_{2} &=28.65 \text { atm } \end{aligned} $$
$$ \begin{array}{l} 2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{N}_{2}+3 \mathrm{H}_{2} \\ (28 .65) \\ 28.65-2 \mathrm{p} \quad p \quad 3 \mathrm{p} \end{array} $$
$$ \begin{aligned} \text { (a) } \varepsilon q^{n} \rightarrow 28.65-2 p+p+3 p &=28.65+2 p \\ 28 \cdot 65+2 p=40.11 \end{aligned} $$
$$ p=5.73 $$
$$ \begin{array}{c} \alpha=\frac{\text { Diss. Pressure }}{\text { Total pressure }}=\frac{2 \times 5.73}{28.65}=0.4 \\ \therefore \alpha=0.4 \end{array} $$
Question 9
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Solid $$Ba{({NO}_{3})}_{2}$$ is gradually dissoved in a $$1\times {10}^{-4}M$$ $${Na}_{2}{CO}_{3}$$ solution. At what minimum conc. of $${Ba}^{-2}$$ will a precipitate of $$Ba{CO}_{3}$$ begin to form? ($${K}_{sp}$$ for $$Ba{CO}_{3}=5.1\times {10}^{-9}$$)

A
$$4.1\times {10}^{-5}M$$

B
$$8.1\times {10}^{-7}M$$

C
$$5.1\times {10}^{-5}M$$

D
$$8.1\times {10}^{-8}M$$

Solution

Alc to the problem
$$\begin{aligned} \mathrm{Na}_{2} \mathrm{CO}_{3} \longrightarrow & 2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \\ \left[\mathrm{cO}_{3}^{2-}\right] &=1 \times 10^{-4} \mathrm{M} \end{aligned} $$
For, $$\mathrm{BaCO}_{3} \rightleftharpoons\left[\mathrm{Ba}^{2+}\right]+\left[\mathrm{CO}_{3}^{2-}\right]$$
$$ \mathrm{KG}_{p}=\left[\mathrm{Ba}^{2+}\right] \cdot\left[\mathrm{CO}_{3}^{2-}\right] $$
$$ \begin{aligned} \Rightarrow \quad 5.1 \times 10^{-9} &=\left[B a^{2+}\right] \cdot\left[1 \times 10^{-4}\right] \\ &\left[\begin{array}{ll} {\left[B a^{2+}\right]=5.1 \times 10^{-9+4}} \\ {\left[B a^{2 +}\right]} =5.1 \times 10^{-5} \mathrm{M} \end{array}\right. \end{aligned} $$
$$\text { (c) is correct option }$$
Question 10
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Consider the following equilibrium system; $$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$$; some inert gas is added to the system at constant volume. Predict which of the following is true?

A
More of $$SO_{3}$$ is produced

B
Less $$SO_{2}$$ is produced

C
Addition of inert gas does not affect equilibrium

D
System moves to new equilibrium position which cannot be predicted theoretically

Solution

When an inert gas is added to an equilibrium system at constant volume, there will be an increase in the total pressure of the system. However, the concentrations of the products and reactants will not change. Therefore, there will be no effect on the equilibrium.

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Equilibrium Test - 56

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Equilibrium Test - 56

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SHARING IS CARING

Question 1

Decreasing the volume of the closed reaction vessel

Decreasing the temperature at which the reaction is performed

Adding an inert gas to the closed reaction vessel

Increasing the volume of the closed reaction vessel

Solution

Question 2

$$B$$ only

Both $$A$$ and $$B$$

Neither $$A$$ nor $$B$$

$$A$$ only

Solution

Question 3

$$2.2\times 10^{-10}$$M

$$1.5\times 10^{-5}$$M

$$1.0\times 10^{-6}$$M

$$5.0\times 10^{-3}$$M

Solution

Question 4

high pressure low temperature

high pressure high temperature

low pressure high temperature

low pressure low temperature

Solution

Question 5

low pressure, high temperature

low pressure, low temperature

high pressure, high temperature

high pressure, low temperature

Solution

Question 6

$$1.5\times 10^{-7}M$$

$$3.0\times 10^{-7}M$$

$$1.5\times 10^{-5}M$$

$$3.0\times 10^{-5}M$$

Solution

Question 7

$$4.2$$

$$4.22$$

$$4.15$$

$$6.4$$

Solution

Question 8

$$0.6$$

$$0.4$$

Unpredictable

None of these

Solution

Question 9

$$4.1\times {10}^{-5}M$$

$$8.1\times {10}^{-7}M$$

$$5.1\times {10}^{-5}M$$

$$8.1\times {10}^{-8}M$$

Solution

Question 10

More of $$SO_{3}$$ is produced

Less $$SO_{2}$$ is produced

Addition of inert gas does not affect equilibrium

System moves to new equilibrium position which cannot be predicted theoretically

Solution

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