Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Thermodynamics question

2024 · 5 Apr · Shift 1 · Q22
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Thermodynamics
  5. /2024 · 5 Apr · Shift 1 · Q22

Thermodynamics question

2024 · 5 Apr · Shift 1 · Q22

JEE MainChemistryThermodynamicsNumerical+4 / −1
The heat of combustion of solid benzoic acid at constant volume is −321.30 kJ-321.30 \mathrm{~kJ}−321.30 kJ at 27∘C27^{\circ} \mathrm{C}27∘C. The heat of combustion at constant pressure is (−321.30−xR) kJ(-321.30-x \mathrm{R}) \mathrm{~kJ}(−321.30−xR) kJ, the value of xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 150

  1. Write the combustion reaction of benzoic acid

Benzoic acid is C6H5COOH\mathrm{C_6H_5COOH}C6​H5​COOH, i.e. C7H6O2\mathrm{C_7H_6O_2}C7​H6​O2​.

Its combustion reaction is:

C7H6O2(s)+152O2(g)→7CO2(g)+3H2O(l)\mathrm{C_7H_6O_2(s) + \frac{15}{2}O_2(g) \rightarrow 7CO_2(g) + 3H_2O(l)}C7​H6​O2​(s)+215​O2​(g)→7CO2​(g)+3H2​O(l)
  1. Use the relation between ΔH\Delta HΔH and ΔU\Delta UΔU

At temperature TTT,

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RTΔH=ΔU+Δng​RT

where Δng\Delta n_gΔng​ is the change in moles of gaseous species.

Here,

  • gaseous products = 777 mol of CO2\mathrm{CO_2}CO2​
  • gaseous reactants = 152\frac{15}{2}215​ mol of O2\mathrm{O_2}O2​

So,

Δng=7−152=−12\Delta n_g = 7 - \frac{15}{2} = -\frac{1}{2}Δng​=7−215​=−21​

Thus,

ΔH=ΔU−12RT\Delta H = \Delta U - \frac{1}{2}RTΔH=ΔU−21​RT
  1. Substitute the given constant-volume heat

Given heat of combustion at constant volume:

ΔU=−321.30 kJ\Delta U = -321.30\,\text{kJ}ΔU=−321.30kJ

Therefore,

ΔH=−321.30−12RT\Delta H = -321.30 - \frac{1}{2}RTΔH=−321.30−21​RT

The question states

ΔH=(−321.30−xR) kJ\Delta H = (-321.30 - xR)\,\text{kJ}ΔH=(−321.30−xR)kJ

Comparing,

xR=12RTxR = \frac{1}{2}RTxR=21​RT

So,

x=T2x = \frac{T}{2}x=2T​

At 27∘C27^\circ \mathrm{C}27∘C,

T=300 KT = 300\,\text{K}T=300K

Hence,

x=3002=150x = \frac{300}{2} = 150x=2300​=150
  1. Final answer
150\boxed{150}150​
  1. Comparison with stored answer

Stored correct answer = 150150150

My derived answer matches the stored answer.

PreviousNext

More from Thermodynamics

  • Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always −57 kJ mol−1…2024 · MCQ
  • Combustion of 1 mole of benzene is expressed at C6​H6​(l)+215​O2​( g)→6CO2​( g)+3H2​O(l).  The standard…2024 · Numerical
  • An ideal gas, Cv​=25​R, is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm…2024 · Numerical
  • For the reaction at 298 K,2 A+B→C,ΔH=400 kJ mol−1 and ΔS=0.2 kJ mol−1 K−1. The reaction will become…2024 · Numerical
  • Consider the figure provided. 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18∘C. If the piston is moved to position B, keeping the temperature unchanged,… Includes diagram2024 · Numerical
  • Δvap ​H⊖ for water is +40.79 kJ mol−1 at 1 bar and 100∘C. Change in internal energy for this vapourisation under same condition is ​kJ mol−1…2024 · Numerical
  • When equal volume of 1 M HCl and 1 M H2​SO4​ are separately neutralised by excess volume of 1MNaOH solution. x and y kJ of heat is liberated…2024 · Numerical
  • The heat of solution of anhydrous CuSO4​ and CuSO4​⋅5H2​O are −70 kJ mol−1 and +12 kJ mol−1 respectively. The heat of hydration of CuSO4​…2024 · Numerical