Thermodynamics Test

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Thermodynamics Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Based on the first law of thermodynamics, which one of the following is correct ?

A
$$\text{For an isochoric process, E = -Q}$$

B
$$\text{For an adiabatic process, E = -Q}$$

C
$$\text{For an isothermal process, E = -Q}$$

D
$$\text{For an cyclic process, W = -Q}$$

Solution

When a system returns to its original state after completing a series of changes, then it is known that a cycle is completed. This process is known as cyclic process. In a cyclic process the initial and the final state is same. As the internal energy U of the system depends only on the state of the system so in a cyclic process the net change of internal energy, ∆U will be equal to zero i.e. ∆U = 0. Hence from the first law (∆U = Q+W)
Hence, W = – Q
Question 2
1 / -0

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements :
v : $${ \Delta H }_{ atomisation }$$ K = 90 kJ/mol
w : $${ \Delta H }_{ ionisation }$$ K = 418 kJ/mol
x : $${ \Delta H }_{ dissociation }$$ H = 436 kJ/mol
y : $${ \Delta H }_{ electron affinity }$$ H = 78 kJ/mol
z : $${ \Delta H }_{ lattice }$$ KH = 710 kJ/mol
In terms of the letters v to z the expression for
(i) $$\Delta $$H for the reaction
$$2K$$(s) + $${ H }_{ 2 }$$(g) $$\longrightarrow $$ $$2KH$$(s).
will be $$\Delta $$$$H =_____?

A
(v + w + x + y + z)

B
(2v + 2w + x + 2y + 2z)

C
(v + w + 2x + y +2 z)

D
(v + 2w + x + y + z)

Solution

In terms of the letters v to z the expression for
(i) $$\Delta $$H for the reaction
$$2K$$(s) + $${ H }_{ 2 }$$(g) $$\longrightarrow $$ $$2KH$$(s).
will be $$\Delta $$$$H = (2v + 2w + x + 2y + 2z)$$.
Question 3
1 / -0

The Born Haber cycle below represents the energy changes occurring at 298K when $$KH$$ is formed from its elements
v : $${ \Delta H }_{ atomisation }$$ $$K = 90 kJ/mol$$
w : $${ \Delta H }_{ ionisation }$$ $$K = 418 kJ/mol$$
x : $${ \Delta H }_{ dissociation }$$ $$H = 436 kJ/mol$$
y : $${ \Delta H }_{ electron affinity }$$ $$H = 78 kJ/mol$$
z : $${ \Delta H }_{ lattice }$$ $$KH = 710 kJ/mol$$
On complete reaction with water, $$0.1 g$$ of $$KH$$ gave a solution requiring 25 $${ cm }^{ 3 }$$ of 0.1M $$HCl$$ for neutralisation.Calculate the relative atomic mass of potassium from this information.

A
$$39$$

B
$$40$$

C
$$41$$

D
None of these

Solution

Meq. of KH = Meq. of $$HCl$$
$$\frac { { 0.1 } }{ { E }_{ KH } } \times 1000\quad =\quad 25\quad \times \quad 0.1$$
Valency factor (05955) of $$K$$ is 1 hence
$${ E }_{ K }$$=$${ M }_{ K }$$ $${ M }_{ K }$$=39
$${ E }_{ KH }$$=40 $${ E }_{ KH }$$=$${ E }_{ K }$$=
40=$${ E }_{ K }$$+1 $${ E }_{ KH }$$
$${ E }_{ K }$$ $$\Rightarrow $$ 39
Question 4
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Directions For Questions

The Born Haber cycle below represents the energy changes occurring at 298K when $$KH$$ is formed from its elements :
v : $${ \Delta H }_{ atomisation }$$ $$K = 90 kJ/mol$$
w : $${ \Delta H }_{ ionisation }$$ $$K = 418 kJ/mol$$
x : $${ \Delta H }_{ dissociation }$$ $$H = 436 kJ/mol$$
y : $${ \Delta H }_{ electron affinity }$$ $$H = 78 kJ/mol$$
z : $${ \Delta H }_{ lattice }$$ $$KH = 710 kJ/mol$$

...view full instructions

Write a balanced equation for the reaction of $$KH$$ with water.

A
$$KH + H_2O \rightarrow KOH + H_2O_2$$

B
$$KH + H_2O \rightarrow KOH + H_2$$

C
$$KH + H_2O \rightarrow K(OH)_2 + H_2$$

D
$$KH + H_2O \rightarrow KOH + H_2O$$

Solution

Potassium hydride hydrolyzes in water to form hydroxide.
$$KH + H_2O \rightarrow KOH + H_2$$
Hence, option B is correct
Question 5
1 / -0

The amount of heat required to raise the temperature of a substance through 1$$^o$$C is called

A
thermal energy

B
calories

C
heat capacity

D
specific heat capacity

Solution

Energy required to raise the temperature of an object by $$1^o C$$ is called heat capacity.
Hence C is correct
Question 6
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Specific heat capacity of lead is $$120\space J\space kg^{-1}\space K^{-1}$$. When $$7200J$$ of heat is supplied to $$5kg$$ of lead, the rise in temperature is

A
$$8^{\small\circ}C$$

B
$$10^{\small\circ}C$$

C
$$12^{\small\circ}C$$

D
$$14^{\small\circ}C$$

Solution

$$Q = ms\triangle t$$
$$\triangle t = \displaystyle\frac{Q}{ms} = \displaystyle\frac{7200}{5\times120} = 12^{\small\circ}$$
Question 7
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$$H_2C=\! = CH_2(g)+ H_2 (g)\longrightarrow H_3C -\!\! \! - CH_3(g)$$
$$The

\: bond \: energy \: of \: C -\!\! \! -H, \:C -\!\! \! -C, \:C=\! =C

\:and \:H-\!\! \! -H \: are \: 414, \: 347,615 \: and \: 435 \: kJ \:

mol^{-1} \: respectively.$$
If the enthalpy change of the above reaction is $$x$$ $$kJ$$, then $$-x$$ is

A
125

B
100

C
225

D
250

Solution

$$\because CH_2=\! =CH_2(g)+ H_2(g)\longrightarrow CH_3-\!\! \! -CH_3(g); \:         \Delta H=?$$
$$\therefore

\Delta H_{Reaction}=Bond \: energy \: data \: used \: for \: formation

\: of \: bond + Bond \: energy \: data \: used \: for \: dissociation \:

of \: bond$$
$$\therefore       \Delta H_{Reaction}= - [e

_{(C-\!\! \! -C)} + 6\times e_{(C-\!\! \! -H)}]+[e_{(C=\! =C)}+4\times

e_{(C-\!\! \! -H)} ]+e_{(H-\!\! \! -H)}$$
                           $$= -e _{(C-\!\! \! -C)} -2\times e_{(C-\!\! \! -H)}+e_{(C=\! =C)}+e_{(H-\!\! \! -H)}$$
                            $$= -347-2\times414+615+435$$
         $$ \Delta H_{Reaction}= -125 \:kJ$$
Question 8
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The $$\Delta H_{f}^{0}$$ for $$CO_{2} (g), CO(g)$$ and $$H_{2}O(g)$$ are $$-393.5, -110.5$$ and $$-241.8\ kJ\ mol^{-1}$$ respectively. The standard enthalpy change (in kJ) for the reaction
$$CO_{2} (g) + H_{2}(g) \rightarrow CO(g) + H_{2}O(g)$$ is

A
$$524.1$$

B
$$41.2$$

C
$$-262.5$$

D
$$-41.2$$

Solution

$$\triangle H = H_{f}^{\circ} (products) - H_{f}^{\circ} (reactants)$$
Reactants are: $$CO_2 \ H_2 \ and \ products\ are \ CO \ and \ H_2O$$
Hence,
$$\Delta H$$ $$=-100.5 + (-241.8) - (-393.5 + 0) = 41.2 \ kJ$$
Hence, option B is correct
Question 9
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Hess's law is used to calculate

A
Enthalpy of reaction

B
Entropy of reaction

C
Work done in reaction

D
All of these

Solution

Correct Option: $$A$$

Explanation:

Hess’s law states that the change in enthalpy in a chemical reaction is the same whether the process is carried out in one step or in multiple steps.

It is the outcome of the first law of thermodynamics. It is used for the calculation of standard reaction enthalpy.

Additional Information:

Hess’s law states that the standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction can be divided, while each occurs at the same temperature.

The enthalpy change for a reaction is independent of the number of ways a product can be obtained if the initial and final conditions are the same.
The negative enthalpy change for a reaction indicates an exothermic process, while positive enthalpy change corresponds to an endothermic process.
Question 10
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For the equilibrium $$H_{2}O(l) \rightleftharpoons H_{2}O(g)$$ at $$1\ atm$$ and $$298\ K$$

A
standard free energy change is equal to zero $$(\triangle G^{\circ} = 0)$$

B
free energy change is less than zero $$(\triangle G < 0)$$

C
standard free energy change is less than zero $$(\triangle G^{\circ} < 0)$$

D
standard free energy change is greater then zero $$(\triangle G^{\circ} > 0)$$

Solution

The process, $$H_{2}O(l) \rightleftharpoons H_{2}O(g)$$,
is an endothermic process, $$(\triangle H = +ve)$$ and entropy increases during this change $$(\triangle S = +ve)$$. Hence this process is spontaneous at all temperatures above $$0^{\circ} C (T\triangle S > \triangle H$$, so $$\triangle G$$ is negative $$\triangle G = \triangle H - T\triangle S)$$. Thus free energy change $$(\triangle G)$$ will be less than zero (negative) at $$1\ atm$$ and $$298\ K$$.

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Thermodynamics Test - 21

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Thermodynamics Test - 21

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Question 1

$$\text{For an isochoric process, E = -Q}$$

$$\text{For an adiabatic process, E = -Q}$$

$$\text{For an isothermal process, E = -Q}$$

$$\text{For an cyclic process, W = -Q}$$

Solution

Question 2

(v + w + x + y + z)

(2v + 2w + x + 2y + 2z)

(v + w + 2x + y +2 z)

(v + 2w + x + y + z)

Solution

Question 3

$$39$$

$$40$$

$$41$$

None of these

Solution

Question 4

$$KH + H_2O \rightarrow KOH + H_2O_2$$

$$KH + H_2O \rightarrow KOH + H_2$$

$$KH + H_2O \rightarrow K(OH)_2 + H_2$$

$$KH + H_2O \rightarrow KOH + H_2O$$

Solution

Question 5

thermal energy

calories

heat capacity

specific heat capacity

Solution

Question 6

$$8^{\small\circ}C$$

$$10^{\small\circ}C$$

$$12^{\small\circ}C$$

$$14^{\small\circ}C$$

Solution

Question 7

125

100

225

250

Solution

Question 8

$$524.1$$

$$41.2$$

$$-262.5$$

$$-41.2$$

Solution

Question 9

Enthalpy of reaction

Entropy of reaction

Work done in reaction

All of these

Solution

Question 10

standard free energy change is equal to zero $$(\triangle G^{\circ} = 0)$$

free energy change is less than zero $$(\triangle G < 0)$$

standard free energy change is less than zero $$(\triangle G^{\circ} < 0)$$

standard free energy change is greater then zero $$(\triangle G^{\circ} > 0)$$

Solution

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