Thermodynamics Test

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Thermodynamics Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

For a particular reaction $$\triangle H^0 = -76.6 KJ$$ and $$\Delta S^0 = 226 JK^{-1}$$. This reaction is:

A
spontaneous at all temperatures

B
non-spontaneous at all temperatures

C
spontaneous at temperature below $$66^0$$C

D
spontaneous at temperature above $$66^0$$C

Solution

Given: Change in enthalpy $$\triangle H^o=-76.6 \ KJ$$
Change in Entropy $$\triangle S^o= 226 \ JK^{-1}$$
Using the relation, $$\triangle G= \triangle H-T \triangle S$$
$$\triangle G = -76.6 \ KJ - T \times 226 \ JK^{-1}$$
Since, both enthalpy $$(\triangle H^o)$$ and Entropy term $$(T \triangle S^o)$$ is negative. Hence $$\triangle G$$ is negative whatever be the temperature.

Question 2

1 / -0

Given the following data:

Substance	$$\Delta H^{o}(kJ/mol)$$	$$\Delta S^{o}(kJ/mol)$$	$$\Delta G^{o}(kJ/mol)$$
$$FeO(s)$$	$$-266.3$$	$$57.49$$	$$-245.12$$
$$C$$(Graphite)	$$0$$	$$5.74$$	$$0$$
$$Fe(s)$$	$$0$$	$$27.28$$	$$0$$
$$CO(g)$$	$$-110.5$$	$$197.6$$	$$-137.15$$

Determine at what temperature the following reaction is spontaneous?
$$FeO(s)+C(Graphite)\rightarrow Fe(s)+CO(g)$$

$$298\ K$$

$$668\ K$$

$$966\ K$$

$$\Delta G^{o}$$ is +ve the reaction will never be spontaneous

Question 3
1 / -0

For the following concentration cell,to be spontaneous $$Pt({H}_{2}){P}_{1} atm |HCl\ || \ Pt({H}_{2}), {P}_{2} atm$$.
Which of the following is correct?

A
$${P}_{1}={P}_{2}$$

B
$${P}_{1}<{P}_{2}$$

C
$${P}_{1}>{P}_{2}$$

D
Can't be predicted

Solution

Solution:- (B) $${P}_{1} < {P}_{2}$$
$${E}_{cell} = {{E}^{0}}_{cell} + \cfrac{0.059}{2} \log{\cfrac{{P}_{2}}{{P}_{1}}}$$
For cell reaction to be spontaneous,
$${E}_{cell} > 0$$
$$\therefore {E}_{cell}$$ will be $$+ve$$ if $${P}_{2} > {P}_{1}$$.
Hence for the cell reaction to be spontaneous,
$${P}_{2} > {P}_{1}$$
Question 4
1 / -0

The change in entropy of $$2$$ moles of an ideal gas upon isothermal expansion at $$243.6K$$ from $$20$$ litre until the pressure becomes $$1atm$$ is:

A
$$1.385\ cal/K$$

B
$$-1.2\ cal/K$$

C
$$1.2\ cal/K$$

D
$$2.77\ cal/K$$

Solution

The change in entropy of 2 moles of an ideal gas upon isothermal expansion at $$243.6\ K$$ from 20 litre until the pressure becomes 1atm is:

Given:
$$P_1 = 1 \,atm$$

No. of moles = 2 mol

T = 243.6 K

V = 20 litre

We know,

$$PV = nRT$$

$$P=\dfrac{nRT}{V}$$

$$P=\dfrac{2\times 0.0821\times 243.6}{20}$$
$$P = 2\, atm$$

$$\Delta S=-4R\,ln\dfrac{P_2}{P_1}$$

$$\Delta S=-4\times \dfrac{8.314}{4.19}\,ln\dfrac{2}{1}$$

$$\Delta S=2\times 0.6931\times 1.9=2.77\,cal/K$$
Change in entropy is $$2.77\, cal/K$$
Question 5
1 / -0

Given the following data:
Substance $$\triangle H^0 (kJ/mol)$$ $$S^0 (J/molK)$$ $$\triangle G^0(kJ/mol)$$
FeO(s) $$-266.3$$ $$57.49$$ $$-245.12$$
C(Graphite) 0 $$5.74$$ 0
Fe(s) 0 $$27.28$$ 0
CO(g) $$-110.5$$ $$197.6$$ $$-137.15$$

Determine at what temperature the following reaction is spontaneous?
$$FeO(s) + C(Graphite) \rightarrow Fe(s) + CO(g) $$

A
$$298$$K

B
$$668$$K

C
$$966$$K

D
$$ \triangle G^0$$ is +ve, hence the reaction will never be spontaneous.

Solution
Question 6
1 / -0

An intimate of ferric oxide $$({Fe}_{2}{O}_{3})$$ and aluminum (Al) is used as solid rocket fuel. Calculate fuel value per gram of the mixture Heats of formation are as follows: $$\triangle { H }_{ f }\left( { Al }_{ 2 }{ O }_{ 3 } \right) =399\quad kcal/mole$$
$$\triangle { H }_{ f }\left( { Fe }_{ 2 }{ O }_{ 3 } \right) =199\quad kcal/mole$$

A
$$0.9345 Kcal/g$$

B
$$200 Kcal/g$$

C
$$3.94Kcal/g$$

D
None

Solution
Question 7
1 / -0

The lattice energy of $$CsI(s)$$ and the enthalpy of solution is $$33\ kJ/mol$$. Calculate the enthalpy of hydration $$(kJ)$$ of 0.65 moles of $$CsI$$.

A
$$738 kJ $$

B
$$ 10 kJ $$

C
$$-371 kJ$$

D
$$-822 kJ$$

Solution
Question 8
1 / -0

Heat of reaction for, $$CO(g)+\dfrac{1}{2}{O}_{2}(g)\rightarrow {CO}_{2}(g)$$ at constant $$V$$ is $$-67.71Kcal$$ at $${17}^{o}C$$. The heat of reaction at constant $$P$$ at $${7}^{o}C$$ is :

A
$$-68.0KCal$$

B
$$+68.0KCal$$

C
$$-67.42Kcal$$

D
None

Solution
Question 9
1 / -0

When two mole of an ideal gas $$\left( { C }_{ p.m }=\cfrac { 5 }{ 2 } R \right) $$ heated from $$300K$$ to $$600K$$ at constant pressure. The change in entropy of gas ($$\Delta S$$) is:

A
$$-\cfrac{3}{2}R\ln{2}$$

B
$$\cfrac{3}{2}R\ln{2}$$

C
$$5R\ln {2}$$

D
$$\cfrac{5}{2}R\ln{2}$$

Solution

Solution:- (C) $$5R \ln{2}$$
$$\Delta{S} = n{C}_{P} \ln{\cfrac{{T}_{2}}{{T}_{1}}} + nR \ln\dfrac{{P}_{1}}{{P}_{2}}$$

$$\because$$ Pressure is constant,

$$\Delta{S} = n {C}_{P} \ln{\cfrac{{T}_{2}}{{T}_{1}}}$$

Given:-
$$n = 2 \text{ moles}$$
$${C}_{P} = \cfrac{5}{2}R$$
$${T}_{2} = 600 K$$
$${T}_{1} = 300 K$$

$$\therefore \Delta{S} = 2 \times \cfrac{5}{2} R \times \ln{\cfrac{600}{300}}$$

$$\Rightarrow \Delta{S} = 5R \ln{2}$$

Hence the change in entropy of the gas is $$5R \ln{2}$$.
Question 10
1 / -0

A system changes its state irreversibly at $$300\ K$$ in which it absorbs $$300\ cals$$ of heat. When the same change is carried out reversibly the amount of heat absorbed is $$900\ cals$$. The change in entropy of the system is equal to?

A
$$1cal$$ $${K}^{-1}$$

B
$$3cal$$ $${K}^{-1}$$

C
$$2cal$$ $${K}^{-1}$$

D
$$1.5cal$$ $${K}^{-1}$$

Solution

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Substance	$$\triangle H^0 (kJ/mol)$$	$$S^0 (J/molK)$$	$$\triangle G^0(kJ/mol)$$
FeO(s)	$$-266.3$$	$$57.49$$	$$-245.12$$
C(Graphite)	0	$$5.74$$	0
Fe(s)	0	$$27.28$$	0
CO(g)	$$-110.5$$	$$197.6$$	$$-137.15$$

Thermodynamics Test - 67

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Thermodynamics Test - 67

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Question 1

spontaneous at all temperatures

non-spontaneous at all temperatures

spontaneous at temperature below $$66^0$$C

spontaneous at temperature above $$66^0$$C

Solution

Question 2

$$298\ K$$

$$668\ K$$

$$966\ K$$

$$\Delta G^{o}$$ is +ve the reaction will never be spontaneous

Question 3

$${P}_{1}={P}_{2}$$

$${P}_{1}<{P}_{2}$$

$${P}_{1}>{P}_{2}$$

Can't be predicted

Solution

Question 4

$$1.385\ cal/K$$

$$-1.2\ cal/K$$

$$1.2\ cal/K$$

$$2.77\ cal/K$$

Solution

Question 5

$$298$$K

$$668$$K

$$966$$K

$$ \triangle G^0$$ is +ve, hence the reaction will never be spontaneous.

Solution

Question 6

$$0.9345 Kcal/g$$

$$200 Kcal/g$$

$$3.94Kcal/g$$

None

Solution

Question 7

$$738 kJ $$

$$ 10 kJ $$

$$-371 kJ$$

$$-822 kJ$$

Solution

Question 8

$$-68.0KCal$$

$$+68.0KCal$$

$$-67.42Kcal$$

None

Solution

Question 9

$$-\cfrac{3}{2}R\ln{2}$$

$$\cfrac{3}{2}R\ln{2}$$

$$5R\ln {2}$$

$$\cfrac{5}{2}R\ln{2}$$

Solution

Question 10

$$1cal$$ $${K}^{-1}$$

$$3cal$$ $${K}^{-1}$$

$$2cal$$ $${K}^{-1}$$

$$1.5cal$$ $${K}^{-1}$$

Solution

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