Thermodynamics Test

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Thermodynamics Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\triangle G$$ for the process $$H_{2}O(s) \rightarrow H_{2}O(l)$$ at $$1\ atm$$ and $$260\ K$$ is:-

A
$$< 0$$

B
$$= 0$$

C
$$> 0$$

D
Unpredictable

Solution
Question 2
1 / -0

The reaction $$NH_{2}CN(s) + \dfrac {3}{2}O_{2}(g) \rightarrow N_{2}(g) + CO_{2}(g) + H_{2}O(l)$$ was carried out at $$300\ k$$ in a bomb calorimeter. The heat released was $$742\ kJ\ mol^{-1}$$. The value of $$\triangle H_{300K}$$ for this reaction would be_________.

A
$$-740.5\ kJ\ mol^{-1}$$

B
$$-741.75\ kJ\ mol^{-1}$$

C
$$-743.0\ kJ\ mol^{-1}$$

D
$$-744.25\ kJ\ mol^{-1}$$

Solution

Enthalpy change for a reaction $$(\Delta H)$$ is given by the expression,
$$\Delta H=\Delta U+\Delta n_gRT$$
where,
$$\Delta U=$$change in internal energy
$$\Delta n_g=$$change in number of moles

For the given reaction,

$$\Delta n_g=\sum{n_g(product}-\sum{n_g(reactant)}$$

$$=(2-1.5)moles$$

$$\Delta n_g=0.5 moles$$,

$$\Delta U=-742.7\ kJ/mol$$

$$T=298\ k$$

$$ R=8.314\times10^{-3} kJ/mol $$

Substituting the values in the expression of $$\Delta H$$

$$\Delta H=(-742.7)+(0.5)(298)(8.314\times10^{-3})$$
$$=-742.7+1.2$$
$$=-741.5\ kJ/mol$$
Question 3
1 / -0

Following reaction occurs at $${25}^{o}C$$:
$$2NO(g, 1\times { 10 }^{ -5 }atm)+{ Cl }_{ 2 }(g, 1\times { 10 }^{ -2 }atm)\rightleftharpoons 2NOCl\ \left( g, 1\times { 10 }^{ -2 }atm \right) $$
$$\Delta { G }^{ o }$$ is_______________.

A
$$-45.65kJ$$

B
$$-28.53kJ$$

C
$$-22.82kJ$$

D
$$-57.06kJ$$

Solution

$$\underset { 1 \times 10^{-5} atm}{2NO}+\underset {1 \times 10^{-2}atm}{Cl_2} \rightleftharpoons \underset { 1 \times 10^{-2} atm}{2NOCl}$$
$$K_p= \cfrac {(P\quad NOCl)^2}{(P\quad NO)^2(P\quad Cl_2)}$$
$$=\cfrac {(1 \times 10^{-2})^2}{(1 \times 10^{-5})^2\times 1 \times 10^{-2}}$$
$$=\cfrac { 1 \times 10^{-2}}{1 \times 10^{-10}}=1 \times 10^8$$
$$\Delta u= -RTln K_p= -8.314 \times 298 \times ln 10^8$$
$$= -8.314 \times 298 \times 8 \times ln10$$
$$= -8.314 \times 298 \times 8 \times 2.3$$
$$= -45.587 J/Kg$$
$$=-45.6 kJ$$ .
Question 4
1 / -0

$$100ml$$ of $${O}_{2}$$ gas diffuses in $$10$$ seconds. $$100ml$$ of gas $$x$$ diffuses in $$t$$ seconds. Gas $$x$$ and time $$t$$ can be respectively:

A
$${H}_{2}, 2.5$$ seconds

B
$${SO}_{2}$$, $$16$$ seconds

C
$$CO$$,$$10$$ seconds

D
$$He$$, $$4$$ seconds

Solution

If the volume of both the gases are same then,
$$\cfrac{t_2}{t_1}=\sqrt{\cfrac{M_2}{M_1}}$$
GIven :
$$t_1=10s$$
$$M_1=32g$$
$$t_2=t$$
$$M_2=M$$
Now, $$\cfrac{t}{10}=\sqrt{\cfrac{M}{32}}$$

Since two unknowns are present we can substitute the values from the given options and then find the right answer.
For $$M=2\ t=2.5sec$$
$$\cfrac{2.5}{10}=\sqrt{\cfrac{2}{32}}$$
$$\cfrac{1}{4}=\sqrt{\cfrac{1}{16}}$$
$$\cfrac{1}{4}=\cfrac{1}{4}$$
LHS=RHS

Hence, Gas x is $$H_2$$ and time t is $$2.5seconds$$
Question 5
1 / -0

When two moles of Hydrogen atoms join together to form a mole of hydrogen molecules in closed rigid vessel with diathermic walls:
$$H(g)+H(g)\longrightarrow { H }_{ 2 }(g)$$

A
$$w< 0$$

B
$$\Delta U=negative$$

C
$${ q }_{ system }=positive\quad $$

D
$$ { q }_{ surrounding }=negative\quad $$

Solution

$$H_{2}$$ has two atoms. So, a mole of $$H_{2}$$ should have 1 mole of $$H_{2}$$ molecules, and when broken down into atoms, 2 moles of atoms.
$$H(g)+H(g)\rightarrow H_{2}(g)$$.
So, from the above reaction $$w<0$$.
Question 6
1 / -0

The difference between the heat of reaction at constant pressure and constant volume for the reaction given below at $${25}^{o}C$$ in KJ is:
$$\quad 2{ C }_{ 6 }{ H }_{ 6(l) }+15{ O }_{ 2(g) }\longrightarrow 12{ CO }_{ 2(g) }+6{ H }_{ 2 }{ O }_{ (l) }$$

A
$$-7.43$$

B
$$+3.72$$

C
$$-3.72$$

D
$$+7.43$$

Solution

The difference between heats of reaction at constant pressure and constant volume for the following reaction at $$25^oC$$ in kJ
$$2C_6H_{6_(l)}+15O_{2_(g)}\longrightarrow12CO_{2_(g)}+6H_2O_{(l)}$$
$$\Delta n=12-15=-3$$
$$\Delta H=\Delta E+RT\Delta n$$
$$\Delta H-\Delta E=RT\Delta n$$
$$=(-3)\times8.314\times10^{-3}\times298$$
$$=-7.443\ kJ/mole$$
Question 7
1 / -0

The normal boiling point of a liquid A is 350 K. $$\Delta { H }_{ vap }$$ at normal boiling point is 35 KJ/mole. Pick out the correct statement(s). (Assume $$\Delta { H }_{ vap }$$ to be independent of pressure).

A
$$\Delta { S }_{ vaporisation }$$ > 100 J/K mole at 350 K and 0.5 atm

B
$$\Delta { S }_{ vaporisation }$$ < 100 J/K mole at 350 K and 0.5 atm

C
$$\Delta { S }_{ vaporisation }$$ < 100 J/K mole at 350 K and 2 atm

D
$$\Delta { S }_{ vaporisation }$$ = 100 J/K mole at 350 K and 2 atm

Solution

The equilibrium at the boiling point is
$$A(1)\leftrightharpoons A(g); \Delta_{vap}H=35\ kJ/mol$$
$$\Delta_{vap}S=\frac{\Delta_{vap}H}{350 K}=100\ J/Kmol$$
At 350 K and 0.5 atm
The boiling point $$(T_{vap})$$ will be lower at the decreased pressure.
Since $$T_{vap}<35K, \Delta_{vap}S>100\ J/Kmol$$
Question 8
1 / -0

Given standard enthalpy of formation of $$CO$$ (-110 KJ $${ mol }^{ -1 }$$) and $$C { O }_{ 2 }$$ (-394 KJ $${ mol }^{ -1 }$$ ). The heat of combustion when one mole of graphite burns is:

A
$$-110\ KJ$$

B
$$-284\ KJ$$

C
$$-394\ KJ$$

D
$$-504\ KJ$$

Solution

Enthalpy of formation of $$CO=-110\, kJ/mol$$
Enthalpy of formtaion of $$CO_{2}=-394\, kJ/mol$$
$$C+O_{2}\rightarrow CO_{2}$$ $$\triangle H_{c}=-394\, kJ/mol$$

Formation of Carbon monoxide
$$C+\cfrac{1}{2}O_{2}\rightarrow CO$$ $$\triangle H_{c}^{'}=-110\, kJ/mol$$

$$\Delta H=\Delta H_{c}- \Delta H_{c}^{'}$$
$$=-394-(-110)$$
$$=-394+110$$
$$=-284\,kJ/mol$$

Hence, the correct option is $$B$$
Question 9
1 / -0

For the reaction, $${ X }_{ 2 }{ O }_{ 4 }\left( l \right) \longrightarrow 2X{ O }_{ 2 }\left( g \right)$$
$$\Delta U=2.1\ k\ cal$$, $$\Delta S=20\ cal\ { K }^{ -1 }$$ at 300 K Hence , $$\Delta G$$ is_________.

A
+2.7 kcal

B
-2.7 kcal

C
+9.3 kcal

D
-9.3 kcal

Solution

The change in Gibbs free energy is given by$$\Delta G=\Delta H -T\Delta S $$
where,
$$\Delta H=$$enthalpy of the reaction
$$\Delta S=$$entropy of reaction

Thus in order to determine $$\Delta G$$, the value of $$\Delta H$$ must be known. The value of $$\Delta H$ can be calculated by the reaction,$$\Delta H=\Delta U+\Delta n_gRT$$
where,
$$\Delta U=$$change in internal energy
$$\Delta n_g=$$(number of moles of gaseous product)-(number of moles of gaseous reactant)$$=2-0=2$$
$$R=$$gas constant$$=2\ cal$$

$$\Delta U=2.1\ kcal=2.1\times10^3\ cal$$
therefore,

$$\Delta H=(2.1\times10^3)+(2\times2\times300)=3300\ cal$$

Hence,
$$\Delta G=\Delta -T\Delta S $$
$$\Delta G=3300 -(300\times20)$$
$$\Delta G=-2700\ cal$$
$$\Delta G=-2.7\ kcal$$
Question 10
1 / -0

The heat liberated on complete combustion of 1 mole of $${ CH }_{ 4 }$$ gas to $${ CO }_{ 2 }\left( g \right)$$ and $${ HO }_{ 2 }\left( l \right)$$ is 890 KJ. Calculate the heat evolved by 2.4 L of $${ CH }_{ 4 }$$ on complete combustion.

A
95.3 KJ

B
8900 KJ

C
89 KJ

D
8.9 KJ

Solution

Given
The volume of $$!\ mol$$ of methane at $$298\ K$$ and $$1\ atm$$ pressure is $$24L$$. Let's calculate the moles of methane for it's $$2.4L$$ as
$$=2.4\left(\frac{1\ mole}{24\ L}\right)$$
$$=0.10\ mol$$
Now, it says that combustion of $$1\ mol$$ of methane generates $$890\ J$$ of heat. We could calculate the heat generated by the combustion of $$0.10\ moles$$ of methane as:
$$=0.10\ mol\left(\frac{890\ kJ}{1\ mole}\right)$$
$$=89kJ$$
Therefore, 89kJ of heat will be evolved by the combustion of 2.4L of methane

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Thermodynamics Test - 63

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Thermodynamics Test - 63

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SHARING IS CARING

Question 1

$$< 0$$

$$= 0$$

$$> 0$$

Unpredictable

Solution

Question 2

$$-740.5\ kJ\ mol^{-1}$$

$$-741.75\ kJ\ mol^{-1}$$

$$-743.0\ kJ\ mol^{-1}$$

$$-744.25\ kJ\ mol^{-1}$$

Solution

Question 3

$$-45.65kJ$$

$$-28.53kJ$$

$$-22.82kJ$$

$$-57.06kJ$$

Solution

Question 4

$${H}_{2}, 2.5$$ seconds

$${SO}_{2}$$, $$16$$ seconds

$$CO$$,$$10$$ seconds

$$He$$, $$4$$ seconds

Solution

Question 5

$$w< 0$$

$$\Delta U=negative$$

$${ q }_{ system }=positive\quad $$

$$ { q }_{ surrounding }=negative\quad $$

Solution

Question 6

$$-7.43$$

$$+3.72$$

$$-3.72$$

$$+7.43$$

Solution

Question 7

$$\Delta { S }_{ vaporisation }$$ > 100 J/K mole at 350 K and 0.5 atm

$$\Delta { S }_{ vaporisation }$$ < 100 J/K mole at 350 K and 0.5 atm

$$\Delta { S }_{ vaporisation }$$ < 100 J/K mole at 350 K and 2 atm

$$\Delta { S }_{ vaporisation }$$ = 100 J/K mole at 350 K and 2 atm

Solution

Question 8

$$-110\ KJ$$

$$-284\ KJ$$

$$-394\ KJ$$

$$-504\ KJ$$

Solution

Question 9

+2.7 kcal

-2.7 kcal

+9.3 kcal

-9.3 kcal

Solution

Question 10

95.3 KJ

8900 KJ

89 KJ

8.9 KJ

Solution

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