Equilibrium Test

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Equilibrium Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

At $$90^oC$$, pure water has $$[H_3O^{\oplus}]=10^{-6.7}\, mol\, L^{-1}$$. What is the value of $$K_w$$ at $$90^oC$$?

A
$$10^{-6}$$

B
$$10^{-12}$$

C
$$10^{-13.4}$$

D
$$10^{-6.7}$$

Solution

At $$90^oC$$, pure water has $$[H_3O^{\oplus}]=10^{-6.7}\, mol\, L^{-1}$$ and $$[OH^-] = 10^{-6.7}$$

So $$K_w = [H^+][OH^-] = 10^{-6.7}\times10^{-6.7} = 10^{-13.4}$$
Question 2
1 / -0

$$pOH$$ of $${H}_{2}O$$ is $$7.0$$ at $$298\ K$$. If water is heated at $$350\ K$$, which of the following should be true?

A
$$pOH$$ will decrease

B
$$pOH$$ will increase

C
$$pOH$$ will remain $$7.0$$

D
Concentration of $${H}^{+}$$ ions will increase but that of $$O{H}^{-}$$ will decrease

Solution

$$K_w = [H^+][OH^-]$$

$$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] ={ 10 }^{ -7 }M$$ for a neutral solution at $$25^o C$$ only.

After that as we increase temp. more no of water molecules dissociate and conc. of both ions increases.

So $$pH$$ and $$pOH$$ both decreases around $$350\:K$$.
Question 3
1 / -0

At $$55^0 C$$, autoprotolysis constant of water is $$4 \times {10}^{-14}$$. If a given sample of water has a $$pH$$ of 6.9, then it is:

A
acidic

B
basic

C
neutral

D
explosive
Question 4
1 / -0

The solubility of $$Ca{ F }_{ 2 } \left( { K }_{ sp } = 3.4 \times { 10 }^{ -11 } \right)$$ in $$0.1 M$$ solution of $$NaF$$ would be

A
$$3.4\times { 10 }^{ -12 }M$$

B
$$3.4\times { 10 }^{ -10 }M$$

C
$$3.4\times { 10 }^{ -9 }M$$

D
$$3.4\times { 10 }^{ -13 }M$$

Solution

$$CaF_2 \rightarrow Ca^{+2} + 2F^- $$
$$S$$ $$0.1$$
$$CaF_2: { K }_{ sp } = 3.4 \times { 10 }^{ -11 } = [Ca^{+2}][F^-]^2$$
$$S = \displaystyle \frac {3.4 \times { 10 }^{ -11 }}{(0.1)^2} = 3.4 \times { 10 }^{ -9 }$$
Question 5
1 / -0

$$C\:(diamond)\rightleftharpoons C\:(graphite)$$ $$\Delta _{r}=-1.9\:kJ/mole$$
Densities of diamond and graphite are $$3.5$$ and $$2.3\:g/mL$$. Favourable conditions for formation of diamond are :

A
high pressure and low temperature

B
low pressure and high temperature

C
high pressure and high temperature

D
low pressure and low temperature

Solution

Conversion from diamond to graphite is an exothermic reaction as $$\Delta_r$$ is negative. Hence, it is an exothermic reaction. So, on increasing the temperature, we are increasing the heat of the reaction. This excess heat can be absorbed in an endothermic reaction. Endothermic reaction takes place during formation of diamond.
Also, during the conversion from diamond to graphite the density of carbon decreases. And density is directly proportional to the molecular mass or the number of moles formed. Hence, number of moles decreases on formation of graphite. If we increase the pressure, density of graphite will decrease. So, the reaction will move such as to increase the density, that is in backward direction.
Thus the reaction takes place at a higher temperatures and higher pressure.
Question 6
1 / -0

The equilibrium $$SO_{2}Cl_{2}(g)\rightleftharpoons SO_{2}(g)+Cl_{2}(g)$$ is attained at $$25^{\circ}C$$ in a closed rigid container when an inert gas, helium is introduced. Which of the following statement is/are correct :

A
concentrations of $$SO_{2},\:Cl_{2}$$ and $$SO_{2}Cl_{2}$$ do not change

B
more chlorine is formed

C
concentration of $$SO_{2}$$ is reduced

D
more $$SO_{2}Cl_{2}$$ is formed

Solution

A closed rigid container is present, so the volume during the reaction remains constant. Now, the reaction constant for concentration involves the moles of the reactant and product sides and the total volume of the container. Inert gases do not take part in the reaction, so the moles remain constant and also the volume is constant. Hence, the reaction constant does change and is same as the equilibrium constant during equilibrium when an inert gas is introduced.
So, the equilibrium concentrations of $$SO_2$$, $$Cl_2$$ and $$SO_2Cl_2$$ do not change on addition of inert gas.
Question 7
1 / -0

An exothermic reaction is represented by the graph:

A

B

C

D

Solution

Hint: Vant hoff’s equation: $$ln \ K_{p} = \dfrac{-\Delta H^{\circ}}{RT} + \dfrac{\Delta S^{\circ}}{R}$$
Explanation:

For an exothermic reaction, $$\Delta H^{\circ} \ is \ negative$$.

Therefore, the equation changes to: $$lnK_{p} = \dfrac{+\Delta H^{\circ}}{RT} + \dfrac{\Delta S^{\circ}}{R}$$

Comparing this equation with the equation of straight line $$y= mx + c$$ , we get
$$y=lnK_{p}$$
$$m=slope= \dfrac{+\Delta H^{\circ}}{R}$$
$$x= \dfrac{1}{T}$$
$$c= \dfrac{\Delta S^{\circ}}{R}$$

Therefore, the slope is positive and constant with a positive intercept.

Hence, option A is correct option.
Question 8
1 / -0

Which of the following expressions is/are not true?

A
$$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] =\sqrt { { K }_{ w } }$$ for a neutral solution at all temperatures

B
$$\left[ { H }^{ + } \right] >\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] <\sqrt { { K }_{ w } } $$ for an acidic solution

C
$$\left[ { H }^{ + } \right] <\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] >\sqrt { { K }_{ w } } $$ for an alkaline solution

D
$$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] ={ 10 }^{ -7 }M$$ for a neutral solution at all temperatures

Solution

$$K_w = [H^+][OH^-]$$

For neutral solution, $$[H^+] = [OH^-]$$
For acidic solution, $$[H^+] > [OH^-]$$
For basic solution, $$[H^+] < [OH^-]$$

So

$$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] =\sqrt { { K }_{ w } }$$ for a neutral solution at all temperatures

$$\left[ { H }^{ + } \right] >\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] <\sqrt { { K }_{ w } } $$ for an acidic solution

$$\left[ { H }^{ + } \right] <\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] >\sqrt { { K }_{ w } } $$ for an alkaline solution

$$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] ={ 10 }^{ -7 }M$$ for a neutral solution at $$25^o C$$ only. After that as we increase temp. more no of water molecules dissociate and conc. of both ions increases.
Question 9
1 / -0

The value of K for a reaction can be changed by changing ____.

A
temperature

B
volume

C
concentration

D
pressure

Solution

True.
The value of equilibrium constant can be altered by varying the temperature.
The variation of $$K$$ with temperature is given by van't Hoff equation.
Question 10
1 / -0

If concentration quotient, $$Q$$ is greater than $${K_c}$$, the net reaction is taking place in backward direction.

A
True

B
False

C
Can not be predicted

D
none of the above

Solution

If concentration quotient, $$Q$$ is greater than $${K_c}$$,the net reaction is taking place in backward direction. This is because, the concentration of products is higher than the equilibrium concentration and the concentration of reactants is lower than the equilibrium concentration. To attain equilibrium, more of the products will be consumed and more of the reactant s will be formed.

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Equilibrium Test - 23

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Equilibrium Test - 23

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Question 1

$$10^{-6}$$

$$10^{-12}$$

$$10^{-13.4}$$

$$10^{-6.7}$$

Solution

Question 2

$$pOH$$ will decrease

$$pOH$$ will increase

$$pOH$$ will remain $$7.0$$

Concentration of $${H}^{+}$$ ions will increase but that of $$O{H}^{-}$$ will decrease

Solution

Question 3

acidic

basic

neutral

explosive

Question 4

$$3.4\times { 10 }^{ -12 }M$$

$$3.4\times { 10 }^{ -10 }M$$

$$3.4\times { 10 }^{ -9 }M$$

$$3.4\times { 10 }^{ -13 }M$$

Solution

Question 5

high pressure and low temperature

low pressure and high temperature

high pressure and high temperature

low pressure and low temperature

Solution

Question 6

concentrations of $$SO_{2},\:Cl_{2}$$ and $$SO_{2}Cl_{2}$$ do not change

more chlorine is formed

concentration of $$SO_{2}$$ is reduced

more $$SO_{2}Cl_{2}$$ is formed

Solution

Question 7

Solution

Question 8

$$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] =\sqrt { { K }_{ w } }$$ for a neutral solution at all temperatures

$$\left[ { H }^{ + } \right] >\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] <\sqrt { { K }_{ w } } $$ for an acidic solution

$$\left[ { H }^{ + } \right] <\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] >\sqrt { { K }_{ w } } $$ for an alkaline solution

$$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] ={ 10 }^{ -7 }M$$ for a neutral solution at all temperatures

Solution

Question 9

temperature

volume

concentration

pressure

Solution

Question 10

True

False

Can not be predicted

none of the above

Solution

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