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Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\text{B}$
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Born-Haber type approach: ΔfH(PCl3) = ΔHatom(P) + 3ΔHatom(Cl) - 3BE(P-Cl). BE(P-Cl) = [314 + 3(121) - 306]/3 = 123.67 kJ/mol. The calculation uses formation enthalpy as the difference between atomization and bond formation energies.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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. $LiCl > LiF > NaF > NaCl$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Lattice enthalpy is inversely proportional to inter-ionic distance. LiF has the smallest ions and thus the highest lattice energy, followed by LiCl, NaF, and NaCl (larger ions).

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

$100 ml$ 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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $0.1N$
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The overall reaction for atomization of OF2 into one oxygen and two fluorine atoms is the sum of the two given steps, so its total enthalpy change is 201 + 199 = 400 kJ. Since OF2 contains two O-F bonds, the average bond energy is obtained by dividing the total enthalpy by two, yielding 400 / 2 = 200 kJ.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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 $-142kJ/mol$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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. $F-H > N-H > O-H$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $N-H$.