Tag: energetics and thermochemistry

Quizzes (6) Questions (104)

Questions Related to energetics and thermochemistry

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

The enthalpies of combustion of carbon and carbon monoxide are -393.5 KJ and -283 KJ respectively the enthalpy of formation of carbon monoxide is :

  1. -676.5 KJ

  2. -110.5 KJ

  3. 110.5 KJ

  4. 676.5 KJ
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The enthalpy of formation of carbon monoxide is calculated using Hess's law by subtracting the enthalpy of combustion of carbon monoxide from that of carbon. Specifically, delta H_f(CO) = delta H_c(C) - delta H_c(CO) = -393.5 - (-283) = -110.5 kJ. This represents the energy change when one mole of carbon monoxide is formed from its elements.

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

The incorrect statement is :

  1. $ \Delta _{ So| }H^{ 0 }={ \Delta } _{ Lattice }H^{ 0+ }\Delta _{ Hyd }H^{ 0 } $

  2. The enthalpy on dilution is independent on original concentration.
  3. $ N \equiv N > C \equiv N > C \equiv C $ [value of mean bond enthalpy in KJ/mol]
  4. $ H _2O(S) \overset { 1\quad bar }{ \rightleftharpoons } H _2O(I) 273 K $


    $ \Delta U= +Ve; \Delta H = +ve; W= +ve; q = +ve $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The statement about dilution enthalpy is incorrect because enthalpy of dilution DOES depend on the original concentration of the solution. Dilution from different starting concentrations releases different amounts of heat. The other statements are correct: Born-Haber cycle relation, bond enthalpy order (N≡N=946 kJ/mol > C≡N=891 > C≡C=619), and ice melting thermodynamics.

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

The enthalpy of tetramerization of $X$ in gas phase $(4X(g)\rightarrow { X } _{ 4 }(g))$ is $-100\ kJ/mol$ at $300\ K$. The enthalpy of vaporisation for liquid $X$ and ${X} _{4}$ are respectively $30\ kJ/mol$ and $72\ kJ/mol$ respectively.
$\Delta S$ for tetramerization of $X$ in liquid phase is $-125\ J/K mol$ at $300\ K$.
What is the $\Delta G$ at $300\ K$ for tetramerization of $X$ in liquid phase?

  1. $-52\ kJ/mol$
  2. $-98\ kJ/mol$
  3. $-14.5\ kJ/mol$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Heat of formation of $2$ moles of ${NH} _{3}(g)$ is $-90kJ$; bond energies of $H-H$ and$N-N$ bonds are $435kJ$ and $390kJ$ ${mol}^{-1}$ respectively. The value of the bond energy of $N\equiv N$ will be:

  1. $-472.5\ kJ$
  2. $-945\ kJ$
  3. $472.5\ kJ$
  4. $945\ kJ$ ${mol}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Delta { H } _{ reaction }=\sum { { \left( BE \right)  } _{ reactants } } -\sum { { \left( BE \right)  } _{ products } } $
$-90=x+3\times 435-6\times 390$
$x=945kJ$ ${mol}^{-1}$

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

The standard enthalpy of formation of ${NH} _{3}$ is $-46kJ{ mol }^{ -1 }$. If the enthalpy of formation of ${H} _{2}$ from its atoms is $-436kJ{ mol }^{ -1 }$ and that of ${N} _{2}$ is $-712kJ{ mol }^{ -1 }$, the average bond enthalpy of $N-H$ bond in ${NH} _{3}$ is:

  1. $+1056kJ{ mol }^{ -1 }\quad $
  2. $-1102kJ{ mol }^{ -1 }\quad $
  3. $-964kJ{ mol }^{ -1 }\quad $
  4. $+352kJ{ mol }^{ -1 }\quad $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ H } _{ reaction }=\sum { { BE } _{ reactants } } -\sum { { BE } _{ products } } $
$-46=(\cfrac(712)+\cfrac{3}{2}(436))-3x$
$x=+352\ kJ{ mol }^{ -1 } $

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

Calculate $ \Delta { H }^{ o }$ of the reaction:

$ { CH } _{ 2 }={ CH } _{ 2 }+3O=O\longrightarrow 2O=C=O+2H-O-H$

The average bond enthelpies of various bond are:

$Bond \quad \quad \quad\quad \quad C-H\quad \quad O=O\quad \quad C=O\quad \quad O-H\quad \quad C=C$
$Bond\ enthalpy$:     414               499               724           460               619
[kJ/mol]

  1. -364kJ

  2. -564kJ

  3. -964kJ

  4. -1654kJ
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using bond enthalpy method: ΔH = Σ(bonds broken) - Σ(bonds formed). Breaking: 4 C-H (4×414=1656), 1 C=C (619), 3 O=O (3×499=1497). Total broken = 3772. Forming: 4 C=O (4×724=2896), 4 O-H (4×460=1840). Total formed = 4736. ΔH = 3772-4736 = -964 kJ. The reaction is exothermic.

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

$H _{2}+Cl _{2}\rightarrow 2HCl+44\ K.Cal$. Heat of decomposition of $HCl$ is:

  1. $-44\ K.Cal$
  2. $+44\ K.Cal$
  3. $-22\ K.Cal$
  4. $+22\ K.Cal$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$H _2+Cl _2 \longrightarrow 2HCl+44Kcal$

Heat of decomposition of $HCl$ :-
$2HCl \longrightarrow \Delta H _{decomp}=+44 Kcal$
$1 HCl \longrightarrow \Delta H _{decomp}=+22 Kcal$
So, Heat of decomposition of $HCl=+22Kcal$

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.