Chemistry

Thermochemistry and Equilibrium

110 Questions

Thermochemistry and equilibrium problems focus on calculating bond energies, lattice enthalpy, and the Born Haber cycle. These concepts are vital for scoring well in chemistry sections. Regular practice ensures a clear understanding of energy changes in reactions.

Bond energy calculationsBorn Haber cycleLattice enthalpyEnthalpy of solutionThermochemical equations

Thermochemistry and Equilibrium Questions

Multiple choice chemistry lattice energy born-haber cycle born-haber cycles energy cycles

The enthalpy of hydrogenation for $1-pentene$ is $+126\ kJ/mol$. The enthalpy of hydrogenation for $1, 3-pentadiene$ is $+230\ kJ/mol$. Hence estimate the resonance magnitude of (delocalization) energy of $1, 3-pentadiene$.

  1. $22\, kJ/mol$
  2. $104\, kJ/mol$
  3. $252\, kJ/mol$
  4. cannot be calculated from this information

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

Enthalpy of hydrogenation per bond$=+126\ kJ/mol$


Enthalpy of hydrogenation for 2 bonds$=+252\ kJ/mol$


Resonance energy$=252-230=22\ kJ/mol$

Multiple choice chemistry lattice energy born-haber cycle born-haber cycles energy cycles

The heats of neutralization of $CH _3COOH.HCOOH, HCN$ and $HClO$ are 13.2, 13.4,2.9 and 3.6 kcal/eq respectively. Then, the degree of hydrolysis for the respective ions will be in the order :

  1. $CH _3COO^- < HCOO^- < CN^- < ClO^-$
  2. $HCOO^- < ClO^- < CN^- < CH _3COO^-$
  3. $CH _3COO^- < CN^- < ClO^- < HCOO^-$
  4. $HCOO^- < CH _3COO^- < ClO^- < CN^-$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Strong acid have high heat of neutralization and are very much stable so required high energy for dissociation. Hence, weak acids are hydrolysed at faster rate than strong acids.

Multiple choice chemistry lattice energy defining lattice energy ionic or electrovalent bond energy cycles

From the following sequence calculate the lattice energy of AB(s):
$A(s)\rightarrow A(g)+e$;      $610\;kJ\;mol^{-1}$
$B(g)+e\rightarrow B(g);$      $-260\;kJ\;mol^{-1}$
$A(s)+B(g)\rightarrow AB(s);$      $-569\;kJ\;mol^{-1}$

  1. $-219$
  2. $-919$
  3. $+1539$
  4. $+301$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The lattice energy of a crystalline solid is usually defined as the energy of formation of the crystal from infinitely-separated ions, molecules, or atoms. 
let the equations be 1, 2 and 3.
We will get the lattice energy by  $3-2-1 $= $-569+260-610$ $= -919$
Multiple choice physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

$\begin{array} { l } { \text { Energy required to dissociate } 4 \mathrm { g } \text { of gaseous } } \ { \text { hydrogen into free gaseous atoms is } 208 \mathrm { Kcal {at}  }  } \ {  25 ^ { \circ } \mathrm { C } \text { . The bond energy of } \mathrm { H } - \mathrm { H } \text { bond will be : } } \end{array}$ .

  1. $1.04Kcal$
  2. $10.4Kcal$
  3. $104Kcal$
  4. $1040Kcal$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given heat of atmosphere $40=260\ Kcal$
$2H _{2} \rightarrow 4H$
$\triangle H = 208\ Kcal$
$20 \rightarrow 1\ mole$
$40 \rightarrow 2 \ mole$
$Hene \ 2 H-H$ bonds area brown $bg $
$20\ kcal $ energy so in order to break $1\ H-H$ bound we required $\dfrac{208}{2}= 104\ Kcal$
Hence the bond energy of $H-H$ bound will be $=104\ kcal$
Multiple choice biology food and human health balanced diet and malnutrition balanced and unbalanced food digestive disorders diseases caused by changes in lifestyle diseases and toxic substance health and its maintenance

Kilocalories of usable energy liberated by one mole of glucose is?

  1. $80$
  2. $160$
  3. $180$
  4. $380$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The aerobic respiration of one mole of glucose typically yields approximately 38 ATP. Given that the hydrolysis of one mole of ATP releases roughly 10 kcal, the total energy is approximately 380 kcal.

Multiple choice chemistry introduction to organic chemistry bonding in organic molecule tetravalence of carbon: shapes of organic compounds introduction to carbon and its compounds

The enthalpy of reaction,${\text{2HC}} \equiv {\text{CH + 5C}}{{\text{O}} _{\text{2}}}{\text{ + 2}}{{\text{H}} _{\text{2}}}{\text{O}}\,$
 If the bounds energies of ${\text{C - H,C}} \equiv {\text{C,}}\,{\text{O = O,C = O}}$ abd ${\text{O - H}}$ bounds are p,q,r,s,t respectively 

  1. $\left[ {8s + 4t} \right] - \left[ {4p + q + 5r} \right]$
  2. $\left[ {4p + 2q + 5r} \right] - \left[ {8s + 4t} \right]$
  3. $\left[ {4p + 2q + 5r + 8s + 4t} \right]$
  4. $\left[ {2q + q + 5r + } \right] - \left[ {8s + 4t} \right]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

Enthalpy of the system is given as :

  1. $\displaystyle H+PV$
  2. $\displaystyle U+PV$
  3. $\displaystyle U-PV$
  4. $\displaystyle H-PV$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Enthalpy: Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been found useful to define a new state function Enthalpy $(H)$ as :
$H = U + PV$ 

Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

Enthalpy of the system is given as

  1. $\,H + PV$
  2. $\,U + PV$
  3. $\,U - PV$
  4. $\,H - PV$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been found useful to define a new state function Enthalpy (H) as :
$H=U+PV $ (By definition)
$\Delta H=\Delta U+\Delta (PV)$
$\Delta H=\Delta U+P\Delta V$ (at constant pressure) combining with first law.
$\Delta H=q _{p}$
Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

If the heat of combustion of carbon monoxide at constant volume and at $17^o$C is $-283.3$ kJ, then its enthalpy of combustion at constant pressure($R=8.314J degree^{-1} mol^{-}$)

  1. $-284.5$ kJ
  2. $284.5$ kJ
  3. $384.5$ kJ
  4. $-384.5$ kJ
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Solution:- (A) $- 284.5 \; kJ$
Heat change at constant volume for the combustion of carbon monoxide $= -283.3 \; kJ$
${CO} _{\left( g \right)} + \cfrac{1}{2} {{O} _{2}} _{\left( g \right)} \longrightarrow {C{O} _{2}} _{\left( g \right)}$
From the above reaction,
$\Delta{{n} _{g}} = {n} _{P} - {n} _{R} = 1 - \left( \cfrac{1}{2} + 1 \right) = - \cfrac{1}{2}$
Temperature $\left( T \right) = 17 ℃ = \left( 17 + 273 \right) K = 290 K \; \left( \text{Given} \right)$
Now from first law of thermodynamics,
$\Delta{H} = \Delta{E} + \Delta{{n} _{g}} RT$
$\Delta{H} = -283.3 + \left( -\cfrac{1}{2} \right) \times 8.314 \times {10}^{-3} \times 290$
$\Rightarrow \Delta{H} = -283.3 - 1.205 = - 284.505 \; kJ$
Hence the heat of reaction at constant pressure will be $- 284.5 \; kJ$.
Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

A sample of $CH _4$ of 0.08 g was subjected to combustion at $27^oC$ in a bomb calorimeter. The temperature of the calorimeter system was found to be raised by $0.25^oC$. If heat capacity of calorimeter is 18 kJ, $\Delta H$ for combustion of $CH _4$ at $27^oC$ is:

  1. $- 900$ kJ/mole
  2. $- 905$ kJ/mole
  3. $- 895$ kJ/mole
  4. $- 890$ kJ/mole
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ CH } _{ 4 }\left( g \right) +{ 2O } _{ 2 }\left( g \right) \longrightarrow { CO } _{ 2 }\left( g \right) +2{ H } _{ 2 }O\left( l \right) $

$\Delta E=$ Heat of combustion
        $=$ Heat capacity $\times$ rise in $T$ $\times$ $\dfrac { Molar\ mass }{ Mass\ of\ compound } $
        $=-18\times 0.25\times \dfrac { 16 }{ 0.08 } $
        $=-900$ KJ/mole                                         $R=8.314\times { 10 }^{ -3 }KJ{ mol }^{ -1 }$
$\Delta H=\Delta E+\Delta nRT$                                      $\Delta n=1-3=-2$
$=-900+\left( -2 \right) \times 8.314\times { 10 }^{ -3 }\times 300$           $T=300K$
$=-900-4.9884=-904.98\simeq -905KJ/mol$

Multiple choice physics energy : forms and sources concept of energy energy for everything forms of energy

Energy equivalent of 4.2 mg in kilo calories is

  1. $9 \times 10^{10} cal$
  2. $8 \times 10^{8} cal$
  3. $10 \times 10^{9} cal$
  4. $6 \times 10^{7} cal$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using E = mc^2. m = 4.2 mg = 4.2 * 10^-6 kg. c = 3 * 10^8 m/s. E = 4.2 * 10^-6 * (9 * 10^16) = 37.8 * 10^10 J. Converting to calories (1 cal = 4.184 J), the result is approximately 9 * 10^10 cal.