Thermodynamics Test

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Thermodynamics Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

For $$Ag, \overline {C_{P}} (JK^{-1}mol^{-1})$$ is given by $$24 + 0.006\ T$$. If temperature of $$3\ mol$$ of silver is raised from $$27^{\circ}C$$ to its melting point $$927^{\circ}C$$ under $$1$$ atm pressure then $$\triangle H$$ is equal to:

A
$$52\ kJ$$

B
$$76.95\ kJ$$

C
$$89.62\ kJ$$

D
$$38.62\ kJ$$

Solution
Question 2
1 / -0

For a certain reaction the change in enthalpy and change in entropy are $$40.63\ kJ\ ,ol\ KJ^{-1}$$. What is the value of $$\triangle G$$ at $$27^{\circ}C$$ and indicate whether the reaction is spontaneous.

A
$$+10630\ J\ mol^{-1}$$; spontaneous

B
$$+10630\ J\ mol^{-1}$$; non spontaneous

C
$$-7990\ J\ mol^{-1}$$; spontaneous

D
$$+7900\ J\ mol^{-1}$$; spontaneous

Solution
Question 3
1 / -0

The heat capacity of a bomb calorimeter is $$300\ J/K$$. When $$0.16\ gm$$ of methane was burnt in this calorimeter the temperature rose by $$3^{\circ}C$$. The value of $$\triangle U$$ per mole will be

A
$$100\ kJ$$

B
$$90\ kJ$$

C
$$900\ kJ$$

D
$$48\ kJ$$

Solution
Question 4
1 / -0

The heat change during the reaction $$24g\ C$$ and $$128g\ S$$ following the change:

$$C + S_{2} \rightarrow CS_{2}; \triangle H = 22K\ cal$$.

A
$$22\ K\ cal$$

B
$$11\ K\ cal$$

C
$$44\ K\ cal$$

D
$$32\ K\ cal$$
Question 5
1 / -0

$$NX$$ is produced by the following step of reactions:
$$M+{ X }_{ 2 }\longrightarrow M{ X }_{ 2 }$$
$$3M{ X }_{ 2 }+{ X }_{ 2 }\longrightarrow { M }_{ 3 }{ X }_{ 8 }$$
$${ M }_{ 3 }{ X }_{ 8 }+{ N }_{ 2 }{ CO }_{ 3 }\longrightarrow NX+{ CO }_{ 2 }+{ M }_{ 3 }{ O }_{ 4 }$$
How much $$M$$ (metal) is consumed to produce $$206gm$$ of $$NX$$?
(Take atomic weight of $$M=56,N=23,X=80$$)

A
$$42gm$$

B
$$336gm$$

C
$$\cfrac { 14 }{ 3 } gm$$

D
$$\cfrac { 7 }{ 4 } gm$$

Solution

$$M+X_{2}\rightarrow MX_{2}$$
$$3MX_{2}+X_{2}\rightarrow M_{3}X_{8}$$
$$M_{3}X_{8}+N_{2}CO_{3}\rightarrow NX+CO_{2}+M_{3}O_{4}$$

Multiplying Equation 1 by Equation 3, and adding all the equations we get
$$3M+3X_{2}\rightarrow 3MX_{2}$$
$$3MX_{2}+X_{2}\rightarrow M_{3}X_{8}$$
$$M_{3}X_{8}+N_{2}CO_{3}\rightarrow NX+CO_{2}+M_{3}O_{4}$$
Resultant equation:- $$3M+4N_{2}+N_{2}CO_{3}\rightarrow NX+CO_{2}+M_{3}O_{4}$$
Mass of $$NX=206\,g$$
Molar mass of $$NX=23+80=103\,g/mol$$
Moles of $$NX=206\,g\times \cfrac{1\,mol}{103\,g}= 2\,mol \,NX$$
Moles of metal $$M= 2\, mol\,NX\times \cfrac{3\,mole\,M}{1\, mole\,NX}=6\,mol\,M$$
Mass of metal $$M=6\,mol\,M\times \cfrac{56\,g\,M}{1 mole\,M}=336\,g\,M$$
Hence, 336g metal is consumed to produce 206g of $$NX$$.
Question 6
1 / -0

Stearic acid $$\left[ C{ H }_{ 3 }\ { \left( C{ H }_{ 2 } \right) }_{ 16 }\ C{ O }_{ 2 }H \right]$$ is a fatty acid, the part of fat that stores most of the energy. $$1.0\ g$$ of stearic acid was burned in a bomb calorimeter. The bomb had a heat capacity of $$652\ J/ C$$. If the temperature of $$500\ g $$ water $$\left( c=4.18J/g\ C \right)$$ rose from $$25.0$$ to $$39.3 C$$, how much heat was released when the strearic acid burned? $$[Given\ { C }_{ p }\left( { H }_{ 0 }O \right) =4.18\ { J }/{ g }\ C]$$

A
$$39.21\ kJ$$

B
$$29.91\ kJ$$

C
$$108\ kJ$$

D
$$9.32\ kJ$$

Solution
Question 7
1 / -0

If an electron is moving in an orbit with total energy $$U = -\dfrac {e^{2}}{r} r$$ is radius of the orbit then find the speed of the electron.

A
$$\dfrac {e}{\sqrt {mr}}$$

B
$$e\dfrac {e}{\sqrt {mr}}$$

C
$$\dfrac {e}{r}\sqrt {\dfrac {2}{m}}$$

D
$$\dfrac {2e}{\sqrt {mr}}$$

Solution
Question 8
1 / -0

$${E}^{o}$$ of an electrode half reaction is related to $$\Delta { G }^{ o }$$ by the equation, $${ E }^{ o }=-\Delta { G }^{ o }/nF$$. If the amount of $${Ag}^{+}$$ in the half reaction, $${ Ag }^{ + }+{ e }^{ - }\longrightarrow Ag$$ is tripled then:

A
$$n$$ is tripled

B
$$\Delta { G }^{ o }$$ reduces to one third

C
$${E}^{o}$$ reduces to one third

D
None of the above

Solution

$$E^{\circ}$$ of an electrode half reaction is related to $$\triangle G^{\circ}$$
by the equation. $$E^{\circ}=\cfrac{-\triangle G^{\circ}}{nF}$$. If the amount of $$Ag^{+}$$ in half reaction. $$Ag^{+}+e\rightarrow Ag$$ is tripled then $$n$$ is tripled.
Question 9
1 / -0

Heat of reaction for the equation, $$A(s) + B(g) \rightarrow 2C(g)$$ is $$40\ kJ$$ at $$300\ K$$ at constant volume. Hence, heat of reaction at constant pressure and at $$300\ K$$ is___________.

A
$$42.5\ kJ$$

B
$$37.5\ kJ$$

C
$$40.0\ kJ$$

D
$$30.0\ kJ$$

Solution
Question 10
1 / -0

On the basis of the following thermochemical data
$${ H }_{ 2 }{ O }_{ (g) }\longrightarrow { H }_{ (aq) }^{ + }+{ OH }_{ (aq) }^{ - };\Delta H=57.32kJ\quad $$
$${H}_{2(g)}+\cfrac{1}{2}{O}_{2(g)}\longrightarrow{ H }_{ 2 }{ O }_{ (l) };\Delta H=-286.2kJ$$
The value of enthalpy of formation of $${OH}^{-}$$ ion at $${ 25 }^{ o }C$$ is:

A
$$+288.88kJ$$

B
$$-343.52kJ$$

C
$$-22.88kJ$$

D
$$-228.88kJ$$

Solution

Solution:- (D) $$-228.88 \; KJ$$
$${{H}_{2}O}_{\left( l \right)} \longrightarrow {{H}^{+}}_{\left( aq. \right)} + {{OH}^{-}}_{\left( aq. \right)}; \quad \Delta{H} = 57.32 \; KJ ..... \left( 1 \right)$$
$${{H}_{2}}_{\left( g \right)} + \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow {{H}_{2}O}_{\left( l \right)}; \quad \Delta{H} = -286.2 \; KJ ..... \left( 2 \right)$$
Adding both the reaction will give enthalpy of formation of $${OH}^{-}$$.
Therefore,
$${{H}_{2}O}_{\left( l \right)} + {{H}_{2}}_{\left( g \right)} + \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow {{H}^{+}}_{\left( aq. \right)} + {{OH}^{-}}_{\left( aq. \right)} + {{H}_{2}O}_{\left( l \right)}; \quad \Delta{H} = -286.2 + 57.32$$
$${{H}_{2}}_{\left( g \right)} + \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow {{H}^{+}}_{\left( aq. \right)} + {{OH}^{-}}_{\left( aq. \right)}; \quad \Delta{H} = -228.88 \; KJ$$
Hence the enthalpy of formation of $${OH}^{-}$$ ion is $$-228.88 \; KJ$$.