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

Calculate the kinetic energy of a gram molecule of argon at 127oC.

  1. 4986 J

  2. 9972 J

  3. 1662 J

  4. 6648 J

  5. 2216 J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Temperature, T = 1270C                               = 273 + 127                               = 400 K R = 8.31 J/mol K Now K.E. per gram molecule of argon = (3/2) x 8.31 x 400 J             = 4986 J

Multiple choice

What is the calorific value of wood?

  1. 33 kJ/g

  2. 17 kJ/g

  3. 3.7 kJ/g

  4. 1.7 kJ/g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Wood has a calorific value of approximately 17 kJ/g, which falls within the typical range of 15-20 kJ/g for dry wood. Option B correctly states this value. Option A (33 kJ/g) is too high and closer to coal. Option C (3.7 kJ/g) is far too low. Option D (1.7 kJ/g) is unrealistically low for wood. The 17 kJ/g value represents the average energy content of good quality dry wood fuel.

Multiple choice

The calorific value of hydrogen is __________.

  1. 32 kJ/g

  2. 77 kJ/g

  3. 150 kJ/g

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Hydrogen has the highest calorific value among common fuels, approximately 150 kJ/g. This is because hydrogen forms water (H₂O) upon combustion, releasing a large amount of energy per gram due to its low atomic mass and strong bond formation.

Multiple choice

Calorific value of L.P.G. is about __________.

  1. 17 megajoules per cubic metre

  2. 55 megajoules per cubic metre

  3. 94 megajoules per cubic metre

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

LPG (Liquefied Petroleum Gas) primarily contains propane and butane. The calorific value of LPG is approximately 88-94 MJ/m³, with 94 MJ/m³ being the standard approximate value. This is significantly higher than natural gas (about 38 MJ/m³), making LPG an efficient fuel.

Multiple choice

Enthalpy change(∆H0) for the reaction is given to be +181 kJmol-1 On increasing the temperature,

  1. more nitric oxide will be formed

  2. nitric oxide will dissociates to give nitrogen and oxygen.

  3. equilibrium will be maintained

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This option is correct and can be explained as follows: Since ∆H0 for the reaction is +ve which means that reaction is endothermic in nature. Hence when temperature is increased then according to Le Chatelier's Principle it favors the forward reaction which nullifies the effect of increase in temperature hence, more nitric oxide is formed.

Multiple choice

Calculate the standard enthalpy of combustion per gram of glucose at 25oC, whereas the standard enthalpies of formation of CO2(g), H2O(I) and glucose at 25oC are -500 kJ/mol, -400 kJ/mol and -1500 kJ/mol, respectively.

  1. -21.66 kJ/g

  2. +21.66 kJ/g

  3. -3900 kJ/g

  4. +3900 kJ/g

  5. +5400 kJ/g
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ∆HoC = Σ ∆Hof (products) - Σ ∆Hof (reactants) = [6 x ∆Hof (CO2) + 6 ∆Hof (H2O)] - [∆Hof (C6H12O6) + 6 x ∆Hof (O2)] = [6 x (-500) + 6 x (-400)] – [-1500 + 6 (0)] = (-3000 - 2400) – (-1500) = -5400 + 1500 = -3900 kJ/mol 
The standard enthalpy of combustion per gram of glucose = -3900/180 = -21.66 kJ/g

Multiple choice methane study of methane carbon- an important element hydrocarbons chemistry

When four moles of $O _2$ is used for complete combustion of methane.$1600kJ$ of heat is evolved.The calorific value of methane is ?

  1. $100 kJ/g$
  2. $800 kJ/g$
  3. $50kJ/g$
  4. $1600 kJ/g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$CH _4+2O _2 \longrightarrow CO _2+2H _2O$             $\Delta H _f$

$\Rightarrow$ for $1$ mole methane   $2$ mole of oxygen is used.
$\Rightarrow$ for $2$ mole methane    $4$ mole of oxygen is used.
$\Rightarrow$ For $2$ mole $\Delta H= 1600$
$\Rightarrow$ For $1$ mole $\Rightarrow \cfrac {\Delta H}{2}=800KJ$
For $1g$ of $CH _4 \Rightarrow \cfrac {800}{16}=50KJ$

Multiple choice chemistry study of the first element - hydrogen occurrence of hydrogen introduction to hydrogen hydrogen

At $298 \, K$ , assuming ideal behaviour, the average kinetic energy of a deuterium molecule is :

  1. Two times that of a hydrogen molecule

  2. Four times that of a hydrogen molecule

  3. Half of that of a hydrogen molecule

  4. Same as that of a hydrogen molecule
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Average Kinetic Energy depends only on temperature

$\displaystyle (KE) _{avg} = \frac {3kT} {2} $ per molecule

Multiple choice modelling gases - the kinetic model ideal gases kinetic theory of gases physics

The kinetic energy of translation of $20gm$ of oxygen at $47^oC$ is (molecule wt. of oxygen is $32gm/mol $and $R=8.3J/mol/K$)

  1. 2490 joules

  2. 2490 ergs

  3. 830 joules

  4. 124.5 joules
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Kinetic energy K = (3/2)nRT. For 20g of oxygen (molar mass 32g/mol), n = 20/32 = 0.625 mol. T = 47 + 273 = 320 K. K = (3/2) * 0.625 * 8.3 * 320 = 2490 J.

Multiple choice chemistry energy fuel cell fuel cells electrochemistry, rechargeable batteries, and fuel cells

If a fuel cell methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is $CH _3OH _{(l)} +\frac{3}{2}O _{2(g)} \rightarrow CO _{2(g)}+2H _2O _{(g)}$ at 298 K standard Gibb's energies of formation for $CH _3OH(l)H _2O(l)$ and $CO _2(g)$ are $-166.2$, $-237.2$ and $394.4$ $KJ \ mol^{-1}$ respectively. If standard enthalpy of combustion of methanol is -726 KJ $mol^{-1}$, efficieny of the fuel cell will be 

  1. 80%

  2. 87%

  3. 90%

  4. 97%
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

The equilibrium constant of the reaction $2C _3H _6 (g) \rightleftharpoons C _2H _4 (g) + C _4H _8 (g)$ is found to fit the expression:
                         $lnK=-1.04-\dfrac {1088}{T}$


Calculate the standard reaction enthalpy and entropy at 400 K :

  1. $\Delta H^o = 4.5 kJ/mol ; \Delta S^o = 4.32 J / mol^{1} K^{1}$
  2. $\Delta H^o = 9.04 kJ/mol ; \Delta S^o =- 8.64 J / mol^{1} K^{1}$
  3. $\Delta H^o = 18.08 kJ/mol ; \Delta S^o = 17.28 J / mol^{1} K^{1}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\displaystyle  \Delta G^0 = -RTlnK =\Delta H^0 - T\Delta S^0$

$\displaystyle  lnK = \frac {\Delta S^0}{R}-\frac {\Delta H^0 }{RT} $......(1) 

But $\displaystyle lnK=-1.04-\frac {1088}{T} $......(2)

From (1) and (2),

$\displaystyle  \frac {\Delta S^0}{R} = -1.04 $ and $\displaystyle \frac {\Delta H^0 }{RT} =  \frac {1088}{T}$

$\displaystyle  \Delta S^0 = -1.04 \times 8.314 = - 8.64 J/mol/K$

$\displaystyle \Delta H^0  = 1088 \times 8.314 =9040 J/mol =9.04 kJ/mol $