Tag: chemistry

Questions Related to chemistry

Two moles of an ideal gas expended isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is:

  1. 11.4

  2. -11.4

  3. 0

  4. 4.8


Correct Option: B
Explanation:
Work done in a reversible isothermal process is:

$W = -2.303 \; nRT \; \log{\cfrac{{V} _{f}}{{V} _{i}}} ..... \left( 1 \right)$

Given:-
$n = 2 \text{ moles}$
$T = 300 \; K$
${V} _{f} = 10 \; L$
${V} _{i} = 1 \; L$
$R =$ Gas constant $= 8.314 \; {J}/{K-mol}$

Substituting these values in ${eq}^{n} \left( 1 \right)$, we have

$W = - 2.303 \times 2 \times 8.314 \times 300 \times \log{\cfrac{10}{1}}$

$\Rightarrow W = -11488.285 \; J = -11.4 \; kJ$

Now as we know that,

$\Delta{H} = \Delta{U} + W$

For an isothermal process,

$\Delta{U} = 0$

$\therefore \Delta{H} = W = -11.4 \; kJ$

Hence the enthalpy change for the given process is $-11.4 \; kJ$.

Hence, the correct option is $\text{B}$

The standard enthalpies of n-pentane, isopentane and neopentane are $-35.0,\ -37.0$ and $-40.0$ $K \ cal/mole$ respectively. The most stable isomer of pentane in terms of energy is ____________.

  1. n-pentane

  2. isopentane

  3. neopentane

  4. n-pentane and isopentane


Correct Option: C
Explanation:

The standard enthalpies of n-pentane, isopentane and neopentane are -35.0, -37.0 and -40.0 K.cal/mole respectively. The most stable isomer of pentane in terms of energy is neopentane as it has most negative value of the standard enthalpy.

Calculate P - CI bond enthalpy 
Given : $\Delta f H(PCl _3, g) = 306 KJ/mol;$     $\Delta H _{atomization} (P, s) = 314 KJ / mol;$
$\Delta f H (Cl, g) = 121 KJ / mol$

  1. 123.66 KJ/mol

  2. 371 KJ / mol

  3. 19 KJ/ mol

  4. None of these


Correct Option: A

Given that bond energies of H-H and Cl-Cl are $430$ and $240$ kJ $mol^{-1}$ respectively and $\Delta _fH$ for HCl is $-90$kJ $mol^{-1}$. Bond enthalpy of HCl is?

  1. $290$ kJ $mol^{-1}$

  2. $380$ kJ $mol^{-1}$

  3. $425$ kJ $mol^{-1}$

  4. $245$ kJ $mol^{-1}$


Correct Option: C

Choose the correct order of lattice enthalpy of $LiCl,\ LiF,\ NaCl$ and $NaF$ :

  1. $LiF > NaCl > NaF > LiCl$

  2. $LiF > LiCl > NaF > NaCl$

  3. $LiF > NaF > Nacl > LiCl$

  4. $LiCl > LiF > NaF > NaCl$


Correct Option: B

$100 ml$ of $0.2\ M\ H _{2}SO _{4}$ is reacted with $100\ ml$ of $0.5\ M\ NaOH$ solution. what is the normality of the solution 

  1. 0.3N

  2. 0.8N

  3. 0.1N

  4. 1N


Correct Option: C
Explanation:
$1M-H _2SO _4=2N-H _2SO _4$

100ml of 0.2M 0.2M $H _2SO _4\equiv 100 \times 0.2$ml of 1M $H _2SO _4$

$\equiv 20ml$ of 2N $H _2SO _4$

$\equiv 40ml$ of 2N $H _2SO _4$

$1M NaOH=1N NaOH$

100ml of 0.2M 0.2M $NaOH\equiv 100 \times 0.2$ml of 1M $NaOH$

$\equiv 20ml$ of 1N $NaOH$

neutralisation occurs when acid and base are mixed due to the formation of salt and water.

20ml of 1N $NaOH\equiv $ 20ml of 1N $H _2SO _4$

$20ml \times 1N=200ml \times $ final strength of acid

therefore the normality of solution is $0.1N$

Calculate the average Bond energy of O-F bond in the following reaction :-
$OF _{2(g)}\rightarrow O _{(g)}+2F _{(g)}$
Given :
$OF _{2(g)}\rightarrow OF _{(g)}+F _{(g)}$; $\Delta H$=201 kJ
$OF _{(g)}\rightarrow O _{(g)}+F _{(g)}$; $\Delta H$=199 kJ

  1. 201 kJ

  2. 199 kJ

  3. 200 kJ

  4. 200.9 kJ


Correct Option: B

Which one of the following statement(s) is/are true?

  1. $\Delta E=0$ for combustion of ${ C } _{ 2 }{ H } _{ 6 }(g)$ in a sealed rigid adiabatic container

  2. ${ \Delta } _{ f }{ H }^{ o }(S,monolithic)\ne 0$

  3. If dissociation energy of $C{ H } _{ 4 }(g)$ is $1656kJ/mol$ and ${ C } _{ 2 }{ H } _{ 6 }(g)$ is $2812kJ/mol$, then value of $C-C$ bond energy will be $328kJ/mol$

  4. If ${ \Delta H } _{ f }({ H } _{ 2 }O,g)=-242kJ/mol; { \Delta H } _{ vap }({ H } _{ 2 }O,l)=44kJ/mol$ then ${ \Delta } _{ f }{ H }^{ o }({{OH}^{-}},aq)$ will be $-142kJ/mol$


Correct Option: A,B,C
Explanation:

${ \Delta } _{ f }{ H }^{ o }(S,monolithic)\ne 0$ as it is not elemental form of S.
$\Delta E=0$ for combustion of ${ C } _{ 2 }{ H } _{ 6 }(g)$ in a sealed rigid adiabatic container as in adiabatic process energy exchange is zero.
For 
${ C } _{ 2 }{ H } _{ 6 }(g)$,
$BE _{C-C} = 2812 - 6\times BE _{C-H} = 2812-6\times 1656/4 = 328$kJ/mol

The bond energy in $kcal : mol^{-1}$ of a $C-! ! ! -C$ single bond is approximately :

  1. 1

  2. 10

  3. 100

  4. 1000


Correct Option: C
Explanation:

The bond energy   of a $C-! ! ! -C$ single bond is approximately 100 $kcal : mol^{-1}$. 


It varies from 82 to $157 :kcal : mol^{-1}$.

Hence, option C is correct.

Bond energies can be obtained by using the following relation:
$\Delta H$(reaction) $=\sum$ Bond energy of bonds, broken in the reactants  $- \sum$ Bond energy of bonds, formed in the products

Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms.

Arrange $N-H$, $O-H$ and $F-H$ bonds in the decreasing order of bond energy:

  1. $F-H > O-H > N-H$

  2. $N-H > O-H > F-H$

  3. $O-H > N-H > F-H$

  4. $F-H > N-H > O-H$


Correct Option: A
Explanation:

Fluorine is more electron-negative than oxygen and oxygen is more electro-negative than nitrogen.

Hence, bond energy between $F-H$ is greater than $O-H$ which is greater than $N-H$.