Thermodynamics Test

Result

Thermodynamics Test - 54

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Compute $$\Delta_rG$$ for the reaction $$H_2O (;, 1 atm, 323 K) \rightarrow H_2O (g, 1 atm, 323 K)$$
Given that : $$\Delta_{vap}H$$ at $$373 K = 40.639 kJmol^{1}, C_P(H_2O, l) = 75.312 J K^{1} mol^{1},
C_P(H_2O, g) = 33.305 J K^{1}mol^{1}$$

A
$$\Delta_rG = 5.59 kJ mol^{1}$$

B
$$\Delta_rG = -5.59 kJ mol^{1}$$

C
$$\Delta_rG = 55.9 kJ mol^{1}$$

D
None of these

Solution

The change in the molar heat capacity at constant pressure for the reaction is as shown.
$$\Delta_rC_P=C_p (H_2O,g) - C_P(H_2O, l) =33.305-75.312=-42.007 J/K mole$$
The entropy change at 323 K is as shown.
$$\displaystyle \Delta_rS_{323}=\frac {\Delta H}{T}=\frac {40639}{323}=108.95 J/K mole$$
The relationship between the entropy change and the change in molar heat capacity at constant pressure is as shown.
$$\displaystyle d(\Delta_rS)=\frac {\Delta_rC_PdT}{T}$$
$$\Delta_rS_{373}-\Delta_rS_{323}=\Delta_rC_P ln\frac {T_2}{T_1}$$
$$\Delta_rS_{373}=108.95-(-42.007 ln \frac {373}{323})$$
$$=115 J/K mole$$
The relationship between the enthalpy change and the change in molar heat capacity at constant pressure is as shown.
$$d(\Delta_rH)=\Delta_rC_PdT$$
$$\Delta_rH_{373} - \Delta_rH_{323} = 42.007 (50)$$
$$\Delta_rH_{373} = 42739.35 J/mole$$
The relationship between Gibbs free energy change, the entropy change and the enthalpy change is as shown.
$$\displaystyle \Delta_rG_{323}= \Delta_rH_{323} - T\Delta_rS_{323}$$
$$\displaystyle \Delta_rG_{323} = 42739.35 -323 (115)$$
$$\displaystyle \Delta_rG_{323} = 5594.35 J = 5.59 kJ/mole$$

Hence, the Gibbs free energy change for the reaction is $$5.59 kJ/mol$$.
Question 2
1 / -0

From the following data of $$\Delta H$$, of the following reaction,
$$C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g);$$ $$\Delta H=-110\:kJ$$
$$C(s)+H_2O(g)\rightarrow CO(g)+H_2(g);$$ $$\Delta H=132\:kJ$$

What is the mole composition of the mixture of oxygen and steam on being passed over coke at 1273K, to keep temperature constant ?

A
0.5 : 1

B
0.6 : 1

C
0.8 : 1

D
1 : 1

Solution

For the system to be in isothermal condition, total enthalpy must be 0.
From energy conservation and thermodynamics,

$$-110\times x\times 2+132\times y = 0$$

$$-220x+132y=0$$

$$y/x = 220/132 = 5:3$$

Hence, $$x:y= 3:5=0.6:1$$

Hence, option B is correct
Question 3
1 / -0

Determine $${ \Delta H }/{ kJ }$$ for the following reaction using the listed enthalpies of reaction:
$$4CO\left( g \right) +8{ H }_{ 2 }\left( g \right) \longrightarrow 3C{ H }_{ 4 }\left( g \right) +C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right) $$

Given that
$$C\left( graphite \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow CO\left( g \right) ; \ { \Delta H }/{ kJ }=-110.5kJ$$

$$CO\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) ;\ { \Delta H }/{ kJ }=-282.9kJ$$

$${ H }_{ 2 }\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( l \right) ;\ { \Delta H }/{ kJ }=-285.8kJ$$

$$C\left( graphite \right) +2{ H }_{ 2 }\left( g \right) \longrightarrow C{ H }_{ 4 }\left( g \right) ;\ { \Delta H }/{ kJ }=-74.8kJ$$

A
$$584.9\:kJ\:mol^{-1}$$

B
$$279.8\:kJ\:mol^{-1}$$

C
$$747.4\:kJ\:mol^{-1}$$

D
$$925\:kJ\:mol^{-1}$$

Solution

$$4CO\left( g \right) +8{ H }_{ 2 }\left( g \right) \longrightarrow 3C{ H }_{ 4 }\left( g \right) +C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right) ;$$ $$\Delta H=?$$

$$C\left( graphite \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow CO\left( g \right) ; \ { \Delta H_1 }=-110.5kJ$$

$$CO\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) ;\ { \Delta H_2 }=-282.9kJ$$

$${ H }_{ 2 }\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( I \right) ;\ { \Delta H_3 }=-285.8kJ$$

$$C\left( graphite \right) +2{ H }_{ 2 }\left( g \right) \longrightarrow C{ H }_{ 4 }\left( g \right) ;\ { \Delta H_4 }=-74.8kJ$$

Applying Hess's law,
$$\Delta H_r =-3\times \Delta H_1+\Delta H_2+2\times \Delta H_3+ 3\times \Delta H_4$$

$$\Delta { H }_{ r } = \left[ 3\left( 110.5 \right) - 282.9 + 2\left( -285.8 \right) + 3\left( -74.8 \right) \right] $$
$$= -747.4$$
Question 4
1 / -0

Fixed amount of an ideal gas contained in a sealed rigid vessel $$(V = 24.6\ litre)$$ at $$1.0$$ bar is heated reversibly from $$27^oC$$ to $$127^oC$$.
Determine change in Gibb's energy (in Joule) if entropy of gas $$S = 10 + 10^{2} T (J/K)$$.

A
$$-530$$ J

B
$$+530$$ J

C
$$530$$ kJ

D
None of the above

Solution

The relationship between the Gibbs free energy, enthalpy and entropy is
$$G = H - TS = U + PV - TS$$
$$dG = dU + PdV + VdP - TdS - SdT$$
$$w = 0, dV = 0, dV = dq = T dS$$
Hence,
$$dG = TdS + VdP - TdS - SdT$$
$$dG = VdP - SdT \Rightarrow \Delta G = V\Delta P-\int S dT$$......(i)
$$VdP = V (P_2 - P_1)$$......(ii)
$$\dfrac {P_2}{T_2}=\dfrac {P_1}{T_1}\Rightarrow \dfrac {P_2}{400}=\dfrac {1}{300}\Rightarrow P_2=\dfrac {4}{3} \times 1 \: bar$$
Substitute value of $$P_2$$ in equation (ii)
$$VdP = 24.6 (4/3- 1) = 8.2 L-atm = 820 J$$......(iii)
$$\int SdT= 2 \displaystyle \int_{T_1}^{T_2} (10+0.01T)dT=10(T_2-T_1)+0.005(T_2^2-T_1^2)$$
$$SdT = 10(100) + 0.005 (400^2 -300^2) = 1350$$.....(iv)
Substitute (iii) and (iv) in (i)
$$\Delta G = 820 -1350 = -530 J$$
Hence, the change in Gibbs energy is $$-530$$ J.
Question 5
1 / -0

Calculate the free energy change at $$298 K$$ for the reaction :
$$Br_2(l) + Cl_2(g) \rightarrow 2BrCl(g)$$. For the reaction $$\Delta H^o = 29.3 kJ$$ & the entropies of
$$Br_2(l), Cl_2(g)$$ & BrCl(g) at the $$298 K$$ are $$152.3, 223.0, 239.7$$ $$J mol^{1} K^{1} $$ respectively

A
$$1721.8 J$$

B
$$1721.8 kJ$$

C
$$17218 J$$

D
None of these

Solution

$$Br_2(l) + Cl_2(g) \rightarrow 2BrCl(g) , \Delta H^o = 29.3 kJ$$
$$ S_{Br} = 152.3 S_{Cl_2(g)} = 223.0$$
$$ S_{BrCl(g)} = 239.7 J mol^{1} K^{1}$$
The entropy change for the reaction is
$$\Delta S_R =2S_{BrCl(g)}- [S_{Br}+S_{Cl_2(g)}]=2 \times 239.7-[ 223.0+ 152.3] = 104.4 \: J mol^{1} K^{1}$$
The relationship between the free energy change, the enthalpy change and the entropy change is as shown.
$$\Delta_rG = \Delta H- T\Delta S$$
Substitute values in the above expression.
$$\Delta_rG = 29300 298 \times 104.4 = 1721.8 J$$
Hence, the free energy change for the reaction is $$1721.8 J$$
Question 6
1 / -0

How many joules of heat are absorbed when 70.0 g of water is completely vaporised at its boiling point ?
[Take : LV = 2260 kJ / kg]

A
22352

B
52460

C
22344

D
158200
Question 7
1 / -0

For the reaction $$X_2O_4(l)\rightarrow 2XO_2(g)$$, $$\Delta U = 2.1\,Kcal,$$ and $$\Delta S = 20\,cal\,K^{ -1 }$$ at 300 K. Hence, $$\Delta G$$ is:

A
$$2.7\,Kcal$$

B
$$-2.7\,Kcal$$

C
$$9.3\,Kcal$$

D
$$-9.3\,Kcal$$

Solution

As we know,

$$\Delta H = \Delta U + \Delta n_gRT$$

$$= 2.1\times 2\times 0.002\times 300 = 3.3\,Kcal$$

$$\Delta G = \Delta H - T\Delta S$$

$$= 3.3\times -300\times (0.02) = -2.7\,Kcal$$

Hence, the correct option is $$\text{B}$$
Question 8
1 / -0

Directions For Questions

Chemist seemingly insatiable need to measure heat has led to a broad spectrum of measurement methods. A special calorimeter is the ice calorimeter, in which the heat released in an exothermic reaction is trapped in an ice -water mixture at 273 K, causing ice to melt which result in contraction of volume.
A student employs this method to determine the heat of combustion of methanol. He finds that when a sample of methanol weighing 0.320g is burnt in excess oxygen in an ice calorimeter at constant volume and 273 K, according to the reaction,

$$2CH_{3}OH(l)+3O_{2}(g)\rightarrow 2CO_{2}(g)+4H_{2}O(l)$$

the volume of ice and water surrounding the sample decreased by 1.82 ml.

[Given: specific volume of ice $$=1.091\:ml/gm$$, specific volume of $$H_{2}O(l)=1\:ml/gm,\:\Delta H_{fusion}$$ of ice$$ =340\:J/gm$$ and $$R=8.314\ J/(mol.K)$$].

...view full instructions

What is internal energy change of combustion for methanol?

A
$$-680\:kJ/mole$$

B
$$-6.8\:kJ/mole$$

C
$$-567\:kJ/mole$$

D
$$-5.67\:kJ/mole$$

Solution

When 1 g of ice melts, the amount of heat absorbed is 340 J. This is as per $$\displaystyle \Delta H _{fusion}$$.

The volume contraction associated with melting of 1 g of ice is $$\displaystyle 1.091-1=0.091$$ ml.

Thus 340 J heat corresponds to 0.091 ml of volume contraction.

The volume contraction of 1.82 ml correspond to $$\displaystyle \frac {340}{0.091} \times 1.82 = 6800$$ J or $$ 6.8 kJ$$.

Since combustion is an exothermic reaction, there is decrease in the internal energy. Due to this, the change in internal energy has negative sign. Hence, the internal energy change of combustion for methanol is$$ - 6.8 kJ/mol$$.
Question 9
1 / -0

Hess's law is used to calculate:

A
Enthalpy of reaction

B
Entropy of reaction

C
Work done in reaction

D
All the above

Solution

Hint: Hess’s law states that, if an overall reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions, at the same temperature.

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.

Extra 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.
Hence from the above explanation, it is clear that it is used to calculate the Enthalpy of reaction
So, the Correct answer is Option A.
Question 10
1 / -0

Given
$$2Fe_2O_3(s) \rightarrow 4Fe(s) +3O_2(g);\Delta G^{\circ}=-1487 KJ mol^{-1}$$

$$2CO(g)+O_2(g) \rightarrow 2CO_2(g);\Delta G^{\circ}=-514.4 KJ mol^{-1}$$

Free energy change $$\Delta G^o$$ for the reaction $$2Fe_2O_3(s) + 6CO(g) \rightarrow 4Fe(s) + 6CO_2(g)$$ will be

A
$$-168.2 kJ\,mol^{-1}$$

B
$$-208.0 kJ\,mol^{-1}$$

C
$$-56.2 kJ\,mol^{-1}$$

D
$$-112.4kJ\,mol^{-1}$$

Solution

1 times the first reaction + 3 times the second reaction gives the desired reaction.
Option C is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 54

Result

Thermodynamics Test - 54

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\Delta_rG = 5.59 kJ mol^{1}$$

$$\Delta_rG = -5.59 kJ mol^{1}$$

$$\Delta_rG = 55.9 kJ mol^{1}$$

None of these

Solution

Question 2

0.5 : 1

0.6 : 1

0.8 : 1

1 : 1

Solution

Question 3

$$584.9\:kJ\:mol^{-1}$$

$$279.8\:kJ\:mol^{-1}$$

$$747.4\:kJ\:mol^{-1}$$

$$925\:kJ\:mol^{-1}$$

Solution

Question 4

$$-530$$ J

$$+530$$ J

$$530$$ kJ

None of the above

Solution

Question 5

$$1721.8 J$$

$$1721.8 kJ$$

$$17218 J$$

None of these

Solution

Question 6

22352

52460

22344

158200

Question 7

$$2.7\,Kcal$$

$$-2.7\,Kcal$$

$$9.3\,Kcal$$

$$-9.3\,Kcal$$

Solution

Question 8

$$-680\:kJ/mole$$

$$-6.8\:kJ/mole$$

$$-567\:kJ/mole$$

$$-5.67\:kJ/mole$$

Solution

Question 9

Enthalpy of reaction

Entropy of reaction

Work done in reaction

All the above

Solution

Question 10

$$-168.2 kJ\,mol^{-1}$$

$$-208.0 kJ\,mol^{-1}$$

$$-56.2 kJ\,mol^{-1}$$

$$-112.4kJ\,mol^{-1}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.