Thermodynamics Test

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Thermodynamics Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

"The change of enthalpy of a chemical reaction is the same whether the reaction takes place in one step or in several steps". This law was presented by:

A
Hess

B
La Chatelier

C
Kirchhoff

D
Lavoisier and Laplace

Solution

According to Hess's law of constant heat summation, even if a chemical reaction takes place by several different routes, the change in enthalpy of the reaction will be same for all the routes.
Question 2
1 / -0

Consider the Born-Haber cycle for the formation of an ionic compound given below and identify the compound $$\left(Z\right)$$ formed.

A
$${M}^{+}{X}^{-}$$

B
$${M}^{-}{X}_{\left(s\right)}^{-}$$

C
$$MX$$

D
$${M}^{+}{X}_{\left(g\right)}^{-}$$

Solution

Based on the above equations it is clear that the cation formed is $$M^{+}$$ and anion formed is $$X^{-}$$. So, the compound formed is $$M^{+}X^{-}$$.
Question 3
1 / -0

If enthalpies of formation for $$C_2H_4(g)$$, $$CO_2(g)$$ and $$H_2O(l)$$ at $$25^o$$C and $$1$$ atm pressure be $$52$$, $$-394$$ and $$-286$$ $$kJ \ mol^{-1}$$ respectively, then the enthalpy of combustion of $$C_2H_4(g)$$ will be:

A
$$-141.2$$kJ $$mol^{-1}$$

B
$$-1412$$kJ $$mol^{-1}$$

C
$$+141.2$$kJ $$mol^{-1}$$

D
$$+1412$$kJ $$mol^{-1}$$

Solution

Given, $$2C(g)+2H_2(g)\rightarrow C_2H_4(g); \Delta H_1=52kJmol^{-1}$$ .......(i)

$$C(g)+O_2(g)\rightarrow CO_2(g); \Delta H_2=-394kJmol^{-1}$$ .......(ii)

$$H_2(g)+\dfrac{1}{2}O_2(g)\rightarrow H_2O(g); \Delta H_3-286kJmol^{-1}$$ ..........(iii)

Reaction involved for combustion of $$C_2H_4(g)$$ is,

$$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O; \Delta H=?$$ .....(iv)

$$2\times[Eq. (ii)+Eq. (iii)]-Eq.(i),$$ we get Eq.(iv)

$$\therefore \Delta =2[\Delta H_2+\Delta H_3]-\Delta H_1$$

$$=2[-394-286]-52$$

$$=-1360 -52=-1412$$ $$kJ\ mol^{-1}$$.

Hence,option B is correct.
Question 4
1 / -0

Spontaneous reactions are :

A
Endergonic

B
Exergonic

C
Energy neutral

D
Exer-endergonic reactions

Solution

Exergonic reactions are the reactions in which energy is released. Initially, exergonic reaction requires activation energy to star the reaction process. Once this activation energy is reached, the reaction proceeds to break bonds and form new bonds. Energy is released as the reaction proceeds. This led to net gain of energy in the surrounding system and a net loss in energy from the reaction system. Hence $$\triangle G$$ is negative and reaction is spontaneous.
Hence we can say that spontaneous reactions are exergonic.
Question 5
1 / -0

For the reactions,
(i) $${H}_{2}(g)+{Cl}_{2}(g)=2HCl(g)+x$$ $$kJ$$

(ii) $${H}_{2}(g)+{Cl}_{2}(g)=2HCl(l)+y$$ $$kJ$$

which one of the following statements is correct?

A
$$x> y$$

B
$$x< y$$

C
$$x-y=0$$

D
$$x=y$$

Solution

(i) $${H}_{2}(g)+{Cl}_{2}(g)=2HCl(g)+x$$ $$kJ$$

(ii) $${H}_{2}(g)+{Cl}_{2}(g)=2HCl(l)+y$$ $$kJ$$

In 2$$^{nd}$$ reaction the product $$HCl$$ formed is in liquid state while in 1$$^{st}$$ reaction it is in gaseous state, so their will be some energy used in convert $$HCl$$ liquid into $$HCl$$ gas due to which the heat evolved in the reaction is less.

Hence, $$x<y$$. the answer is option $$B$$.
Question 6
1 / -0

A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If $$\Delta$$H is the enthalpy change and $$\Delta$$E is the change in internal energy, then:

A
$$\Delta H > \Delta E$$.

B
$$\Delta H < \Delta E$$.

C
$$\Delta H = \Delta E$$.

D
the relationship depends on the capacity of the vesssel.

Solution

$$\displaystyle 2CO +O_2\rightarrow 2CO_2$$

$$\displaystyle \Delta n$$ = no. of moles of gaseous product$$-$$no. of moles of gaseous reactant
$$\displaystyle \Delta n = 2 - (2+1) = -1$$

$$\displaystyle \Delta H = \Delta E + \Delta nRT $$

$$\displaystyle \Delta H = \Delta E + (-1)RT $$

$$\displaystyle \Delta H = \Delta E - RT $$

$$\displaystyle \Delta H < \Delta E $$
Hence,option B is correct.
Question 7
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The enthalpies of combustion of $${C}_{(graphite)}$$ and $${C}_{(diamond)}$$ are $$-393.5$$ and $$-395.4kJ/mol$$ respectively. The enthalpy of conversion of $${C}_{(graphite)}$$ to $${C}_{(diamond)}$$ in $$kJ/mol$$ is:

A
$$-1.9$$

B
$$-788.9$$

C
$$1.9$$

D
$$788.9$$

Solution

The combustion reactions for $$C_{(graphite)}$$ and $$C_{(diamond)}$$ can be written as,
$$C_{(graphite)} + O_2 \rightarrow CO_2$$ $$\Delta H_1 = -393.5 \dfrac{KJ}{mol}$$ (1)
$$C_{(diamond)} + O_2 \rightarrow CO_2$$ $$\Delta H_2 = -395.4 \dfrac{KJ}{mol}$$ (2)
On subtracting reaction (2) from reaction (1) we will get,
$$C_{(graphite)} \rightarrow C_{(diamond)}$$ with $$\Delta H = \Delta H_1 - \Delta H_2 = -393.5 - (-395.4) = 1.9\dfrac{KJ}{mol}$$
Hence, answer is option C.
Question 8
1 / -0

If $$\Delta {H}_{f}^{o}$$ for $${H}_{2}{O}_{2}(l)$$ and $${H}_{2}O(l)$$ are $$-188kJ$$ $${mol}^{-1}$$ and $$-286kJ$$ $${mol}^{-1}$$, what will be the enthalpy change of the reaction?
$$2{H}_{2}{O}_{2}(l)\rightarrow 2{H}_{2}O(l)+{O}_{2}(g)$$

A
$$146kJ$$ $${mol}^{-1}$$

B
$$-196kJ$$ $${mol}^{-1}$$

C
$$-494kJ$$ $${mol}^{-1}$$

D
$$-98kJ$$ $${mol}^{-1}$$

Solution

For the given reaction,
$$2H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)$$
enthalpy change of the reaction = $$\Delta H $$ = $$2\times \Delta H_{f(H_2O)} - 2\times \Delta H_{f(H_2O_2)}$$ (since, $$\Delta H_{f(O_2)} = 0$$ )
hence, $$\Delta H = 2\times (-286) -2\times (-188) = -196KJ\text{ mol}^{-1}$$
Hence, answer is option $$B$$.
Question 9
1 / -0

From the thermochemical reactions,
$${C}_{(graphite)}+\cfrac{1}{2}{O}_{2}\rightarrow CO;\Delta H=-110.5kJ$$
$$CO+\cfrac{1}{2}{O}_{2}\rightarrow {CO}_{2};\Delta H=-283.2kJ$$
the heat of reaction of $${C}_{(graphite)}+{O}_{2}\rightarrow {CO}_{2}$$ is:

A
$$+393.7kJ$$

B
$$-393.7kJ$$

C
$$-172.7kJ$$

D
$$+172.7kJ$$

Solution

Given reactions are,
$$C_{(graphite)} +\dfrac{1}{2}O_2 \rightarrow CO$$ $$ \Delta H = -110.5KJ$$
$$CO + \dfrac{1}{2}O_2 \rightarrow CO_2$$ $$ \Delta H = -283.2KJ$$
Since on adding both the above reaction we will get a final reaction
$$C_{(graphite)} +O_2 \rightarrow CO_2$$
which is the same equation whose heat of reaction is being asked.
Since, on adding two reactions the heat of reactions gets added.
Hence, the heat of reaction for the final reaction will be,
$$\Delta H = \Delta H_1 + \Delta H_2 = -110.5 - 283.2 = 393.7KJ$$
Hence, answer is option B.
Question 10
1 / -0

Given:
$$C(s)+{O}_{2}(g)\rightarrow {CO}_{2}(g)+94.2kcal$$
$${H}_{2}(g)+\cfrac{1}{2}{O}_{2}(g)\rightarrow {H}_{2}O(l)+68.3kcal$$
$${CH}_{4}+2{O}_{2}(g)\rightarrow {CO}_{2}(g)+2{H}_{2}O(l)+210.8kcal$$
The heat of formation of methane in $$kcal$$ will be:

A
$$45.9$$

B
$$47.8$$

C
$$20.0$$

D
$$47.3$$

Solution

First reaction is formation reaction of carbon dioxide. Hence, $$\text{heat of formation of carbon dioxide = 94.2Kcal.}$$
Second reaction is formation reaction of water. Hence, $$\text{heat of formation of water = 68.3kCal.}$$
Since, $$\Delta H$$ for third reaction = $$210.8Kcal$$
Also, $$\Delta H $$ = $$\Delta H_{f(CO_2)} + 2\times \Delta H_{f(H_2O)} - \Delta H_{f(CH_4)}$$
=> $$210.8 = 94.2 + 2\times 68.3 - \Delta H_{f(CH_4)}$$
=> $$\Delta H_{f(CH_4)} = 94.2 +136.6 -210.8 = 20Kcal.$$
Hence, answer is option $$C$$.

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Thermodynamics Test - 27

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Thermodynamics Test - 27

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Question 1

Hess

La Chatelier

Kirchhoff

Lavoisier and Laplace

Solution

Question 2

$${M}^{+}{X}^{-}$$

$${M}^{-}{X}_{\left(s\right)}^{-}$$

$$MX$$

$${M}^{+}{X}_{\left(g\right)}^{-}$$

Solution

Question 3

$$-141.2$$kJ $$mol^{-1}$$

$$-1412$$kJ $$mol^{-1}$$

$$+141.2$$kJ $$mol^{-1}$$

$$+1412$$kJ $$mol^{-1}$$

Solution

Question 4

Endergonic

Exergonic

Energy neutral

Exer-endergonic reactions

Solution

Question 5

$$x> y$$

$$x< y$$

$$x-y=0$$

$$x=y$$

Solution

Question 6

$$\Delta H > \Delta E$$.

$$\Delta H < \Delta E$$.

$$\Delta H = \Delta E$$.

the relationship depends on the capacity of the vesssel.

Solution

Question 7

$$-1.9$$

$$-788.9$$

$$1.9$$

$$788.9$$

Solution

Question 8

$$146kJ$$ $${mol}^{-1}$$

$$-196kJ$$ $${mol}^{-1}$$

$$-494kJ$$ $${mol}^{-1}$$

$$-98kJ$$ $${mol}^{-1}$$

Solution

Question 9

$$+393.7kJ$$

$$-393.7kJ$$

$$-172.7kJ$$

$$+172.7kJ$$

Solution

Question 10

$$45.9$$

$$47.8$$

$$20.0$$

$$47.3$$

Solution

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