Thermodynamics Test

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Thermodynamics Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

A sample of liquid in a thermally insulated container (a calorimeter) is stirred for 2 hr by a mechanical linkage to a motor in the surrounding. For this process:

A
$$w < 0; q = 0; \Delta U = 0$$

B
$$w > 0; q < 0; \Delta U > 0$$

C
$$w < 0; q > 0; \Delta U = 0$$

D
$$w > 0; q = 0; \Delta U > 0$$

Solution
Question 2
1 / -0

A certain amount of potassium chlorate on thermal decomposition gives oxygen which is sufficient for the combustion of ethane. When the products are cooled, the volume of the gaseous product is v ml. Identify the correct sequence of steps for the calculation of the mass of potassium chlorate
(a) Calculation of oxygen required for the combustion of ethane.
(b) Calculation of the amount of the ethane subjected to combustion from the volume of gaseous product.
(c) Calculation of potassium chlorate which can give the required amount of oxygen.
(d) Identification of the products of combustion of ethane and the product left after the cooling of products.

A
a b c d

B
d b a c

C
b c a d

D
d a c b

Solution

Thermal decomposition of $$KCl{O}_{3}$$
$$2KCl{ O }_{ 3 }\left( s \right) \rightarrow 2KCl\left( s \right) +3{ O }_{ 2 }\left( g \right) $$
Combustion of ethane $${ C }_{ 2 }{ H }_{ 6 }\left( g \right) +\cfrac { 7 }{ 2 } { O }_{ 2 }\left( g \right) \rightarrow 2C{ O }_{ 2 }\left( g \right) +3{ H }_{ 2 }O$$
To find the mass of potassium chlorate
First we will identify the products of combustion of ethane and the products left after cooling of products
We required the products left after cooling so that the mass of the product is measured correct
From the volume of gases formed in the combustion of ethane, we can find the mass of ethane
From the mass of ethane we can find the amount of oxygen which is required to form products
Since the oxygen required for combustion of ethane is formed from the thermal decomposition of $$KCl{O}_{3}$$. So the amount of oxygen used in combustion of ethane is the same which is formed from the thermal decomposition of $$KCl{O}_{3}$$.
From the amount of oxygen, we can find the mass of potassium chlorate.
Question 3
1 / -0

Calculate $${ \Delta H }/{ kJ }$$ for the following reaction using the listed standard enthalpy of reaction data:
$$2{ N }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right) $$

$${ N }_{ 2 }\left( g \right) +3{ O }_{ 2 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right)   { \Delta H }/{ kJ }=-414.0$$

$${ N }_{ 2 }{ O }_{ 5 }\left( s \right) +{ H }_{ 2 }O\left( I \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right)   { \Delta H }/{ kJ }=-86.0$$

$$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( I \right)   { \Delta H }/{ kJ }=-571.6$$

A
$$-84.4$$

B
$$-243.6$$

C
$$-71.2$$

D
$$-121.8$$

Solution

For the reaction, $$2{ N }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right) $$

$${ N }_{ 2 }\left( g \right) +3{ O }_{ 2 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right)   { \Delta H }/{ kJ }=-414.0$$

$${ N }_{ 2 }{ O }_{ 5 }\left( s \right) +{ H }_{ 2 }O\left( I \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right)   { \Delta H }/{ kJ }=-86.0$$

$$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( I \right)   { \Delta H }/{ kJ }=-571.6$$

Applying Hess's law,

$$\Delta { H }_{ r }=\left[ 2\left( -414 \right) +2\left( 86 \right) +571.6 \right] $$

$$\Delta { H }_{ r }=-84.4 kJ$$
Question 4
1 / -0

The $$\Delta_f H^{ \ominus }$$ for $$CO_2(g), CO(g) and H_2O(g)$$ are $$-393.5, -110.5 and -214.8 kJ mol^{ -1 }$$, respectively. The standard enthalpy change $$(in kJ mol^{ -1 })$$ for the reaction $$CO_2(g) + H_2(g)\longrightarrow CO(g) + H_2O(g)$$ is:

A
524.1

B
+41.2

C
-262.5

D
-41.2

Solution

$$CO_2(g) + H_2(g)\longrightarrow CO(g) + H_2O(g)$$
$$\Delta H^{

\ominus } = \left\{ \left[ \Delta _{ f }H^{ \ominus }CO,(g)+\Delta _{ f

}H^{ \ominus }H_{ 2 }O,(g) \right] -\Delta _{ f }H^{ \ominus }CO_{ 2

},(g) \right\}$$
$$\quad\quad = \left[ (-110.5)+(-241.8)-(-393.5+0) \right] $$
$$\quad\quad = +41.2 kJ$$
$$\Delta_fH^{ \ominus }H_2, (g) = 0$$
$$\Delta_fH^{ \ominus } = 0$$ in elementary state.
Question 5
1 / -0

$$H_2(g) + \frac{ 1 }{ 2 }O_2(g)\longrightarrow 2H_2O(l); \Delta H = -86 kJ$$
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l)......... kJ(\pm?)$$

A
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 172 kJ$$.

B
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) - 172 kJ$$.

C
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) +0 kJ$$.

D
None of these

Solution

Reaction:
$$H_2(g) + \frac{ 1 }{ 2 }O_2(g)\longrightarrow 2H_2O(l); \Delta H = -86 kJ$$.................1
multiply it by 2 we will get,
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l); \Delta H = -86*2 = -172 kJ$$
which can be written as
$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 172 kJ$$.
Question 6
1 / -0

According to the given reaction, when a $$45 $$ gram sample of the ethanol is burned with excess oxygen, how much energy is released in the form of heat?

$$C_2H_5OH(l) + 3O_2(g)\rightarrow 2CO_2(g) + 3H_2O(l)$$

$$\triangle H = -1.40\times 10^3$$ kJ.

A
0.995 kJ

B
$$5.1 \times 10^2 $$ kJ

C
$$1.40 \times 10^3$$ kJ

D
$$2.80 \times 10^3$$ kJ

E
5000 kJ

Solution
Question 7
1 / -0

The temperature in K at which $$\displaystyle \Delta G=0$$, for a given reaction with $$\displaystyle \Delta H=-20.5\ kJ{ mol }^{ -1 }$$ and $$\displaystyle \Delta S=-50.0\ J{ K }^{ -1 }{ mol }^{ -1 }$$ is:

A
-410

B
410

C
2.44

D
-2.44

Solution

Given that,
$$\displaystyle \Delta H = - 20.5\ kJ/mol = -20500\ J/mol$$

$$\Delta S=-50\ JK^{-1}mol^{-1}$$

Now,

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

$$\displaystyle 0 = -20500 - T (-50.0)$$

$$\displaystyle 20500 = 50T$$

$$\displaystyle \Rightarrow T = 410\ K$$
Question 8
1 / -0

If 10 g of liquid at 300K is heated to 350 K, the liquid absorbs 6 kcals. Determine the specific heat of the liquid in cal/$$\displaystyle { g }^{ \circ }C$$.

A
6

B
12

C
60

D
120

E
600

Solution

Specific heat = $$\dfrac {energy\space absorbed}{sample\space mass\times Temperature\space change} =\dfrac {6000} {10\times 50}= 12 cal/g$$
Question 9
1 / -0

How much energy is released when $$6\ moles$$ of octane is burnt in air? Given $$\triangle H_{f}^{\circ}$$ for $$CO_{2} (g), H_{2}O(g)$$ and $$C_{8}H_{18} (l)$$ respectively are $$-490, -240$$ and $$+160\ J/mol$$

A
$$-6.2\ kJ$$

B
$$-37.4\ kJ$$

C
$$-35.5\ kJ$$

D
$$-20.0\ kJ$$

Solution

$$C + O_{2} = CO_{2} \triangle H = -490 .... (i)$$
$$H_{2} + \dfrac {1}{2} O_{2} = C_{8} H_{18} \triangle H = -240 ...... (ii)$$
$$8C + 9H_{2} = C_{8} H_{18} \triangle H = +160 ...... (iii)$$
$$2 C_{8} H_{18} + 25O_{2} = 16CO_{2} + 18H_{2}O$$
The required reaction can be obtained by
$$2\times (iii) - 16 (i) - 18(ii)$$
$$16C + 18H_{2} - 16C - 16O_{2} - 18H_{2} - 9O_{2}$$
$$= 2C_{8}H_{18} - 16CO_{2} - 18H_{2}O$$
or $$-25O_{2} = 2C_{8} H_{18} - 16CO_{2} - 18H_{2}O$$
$$2C_{8}H_{18} + 25O_{2} = 16CO_{2} + 18H_{2}O$$
$$[\triangle H = 2\times 160 - 16\times -490 - 18 \times -240]$$
$$\triangle H = -11840\ J = -11840\ kJ$$ for $$2\ moles$$ of octane. Energy released for $$6\ moles$$ of octane $$=-11.840 \times 3 = -35.5\ kJ$$.
Hence, option C is correct
Question 10
1 / -0

Bond Average Bond Energy (kJ/mol)
$$C \equiv O$$ 1075
$$C=O$$ 728
$$C-Cl$$ 326
$$Cl-Cl$$ 243
From the above given data, calculate the heat of reaction for the following reaction :

$$CO+{ Cl }_{ 2 }\rightarrow CO{ Cl }_{ 2 }$$

A
$$+62$$ kJ

B
$$-62$$ kJ

C
$$-409$$ kJ

D
$$+706$$ kJ

Solution

$$CO+Cl_2 \rightarrow COCl_2$$

Heat of reaction = Sum of bond Energy of products - sum of bond energy of reactant.

Heat of reaction = $$[728 + 2\times 326 ] - [1075 + 243] = 1380 - 1318 = +62\ kJ$$

Hence, the correct option is A.

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Bond	Average Bond Energy (kJ/mol)
$$C \equiv O$$	1075
$$C=O$$	728
$$C-Cl$$	326
$$Cl-Cl$$	243

Thermodynamics Test - 55

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Thermodynamics Test - 55

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Question 1

$$w < 0; q = 0; \Delta U = 0$$

$$w > 0; q < 0; \Delta U > 0$$

$$w < 0; q > 0; \Delta U = 0$$

$$w > 0; q = 0; \Delta U > 0$$

Solution

Question 2

a b c d

d b a c

b c a d

d a c b

Solution

Question 3

$$-84.4$$

$$-243.6$$

$$-71.2$$

$$-121.8$$

Solution

Question 4

524.1

+41.2

-262.5

-41.2

Solution

Question 5

$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 172 kJ$$.

$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) - 172 kJ$$.

$$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) +0 kJ$$.

None of these

Solution

Question 6

0.995 kJ

$$5.1 \times 10^2 $$ kJ

$$1.40 \times 10^3$$ kJ

$$2.80 \times 10^3$$ kJ

5000 kJ

Solution

Question 7

-410

410

2.44

-2.44

Solution

Question 8

6

12

60

120

600

Solution

Question 9

$$-6.2\ kJ$$

$$-37.4\ kJ$$

$$-35.5\ kJ$$

$$-20.0\ kJ$$

Solution

Question 10

$$+62$$ kJ

$$-62$$ kJ

$$-409$$ kJ

$$+706$$ kJ

Solution

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