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Thermodynamics question

2024 · 5 Apr · Shift 2 · Q28
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  5. /2024 · 5 Apr · Shift 2 · Q28

Thermodynamics question

2024 · 5 Apr · Shift 2 · Q28

JEE MainChemistryThermodynamicsNumerical+4 / −1
Combustion of 1 mole of benzene is expressed at C6H6(l)+152O2( g)→6CO2( g)+3H2O(l). \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \text {. }C6​H6​(l)+215​O2​( g)→6CO2​( g)+3H2​O(l).  The standard enthalpy of combustion of 2 mol2 \mathrm{~mol}2 mol of benzene is −′x′kJ-^{\prime} x^{\prime} \mathrm{kJ}−′x′kJ. x=x=x=‾\underline{\hspace{2cm}}​. Given : 1. standard Enthalpy of formation of 1 mol1 \mathrm{~mol}1 mol of C6H6(l)\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})C6​H6​(l), for the reaction 6C6 \mathrm{C}6C (graphite) +3H2(g)→C6H6(l)+3 \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+3H2​(g)→C6​H6​(l) is 48.5 kJ mol−148.5 \mathrm{~kJ} \mathrm{~mol}^{-1}48.5 kJ mol−1. 2. Standard Enthalpy of formation of 1 mol1 \mathrm{~mol}1 mol of CO2(g)\mathrm{CO}_2(\mathrm{g})CO2​(g), for the reaction C\mathrm{C}C (graphite) +O2(g)→CO2(g)+\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})+O2​(g)→CO2​(g) is −393.5 kJ mol−1-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}−393.5 kJ mol−1. 3. Standard and Enthalpy of formation of 1 mol1 \mathrm{~mol}1 mol of H2O(l)\mathrm{H}_2 \mathrm{O}(\mathrm{l})H2​O(l), for the reaction H2(g)+12O2(g)→H2O(l)\mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})H2​(g)+21​O2​(g)→H2​O(l) is −286 kJ mol−1-286 \mathrm{~kJ} \mathrm{~mol}^{-1}−286 kJ mol−1.
Numerical answer
View written solutionFree

Correct answer: 6535

  1. Use Hess's law

For the combustion reaction of 1 mole benzene:

C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)\mathrm{C_6H_6(l)}+\frac{15}{2}\mathrm{O_2(g)} \rightarrow 6\mathrm{CO_2(g)}+3\mathrm{H_2O(l)}C6​H6​(l)+215​O2​(g)→6CO2​(g)+3H2​O(l)

The standard enthalpy change is:

ΔHcomb∘=∑nΔHf∘(products)−∑nΔHf∘(reactants)\Delta H^\circ_{\text{comb}} = \sum n\Delta H_f^\circ(\text{products}) - \sum n\Delta H_f^\circ(\text{reactants})ΔHcomb∘​=∑nΔHf∘​(products)−∑nΔHf∘​(reactants)

Since ΔHf∘(O2)=0\Delta H_f^\circ(\mathrm{O_2})=0ΔHf∘​(O2​)=0,

ΔHcomb∘=[6ΔHf∘(CO2)+3ΔHf∘(H2O)]−ΔHf∘(C6H6)\Delta H^\circ_{\text{comb}} = \left[6\Delta H_f^\circ(\mathrm{CO_2}) + 3\Delta H_f^\circ(\mathrm{H_2O})\right] - \Delta H_f^\circ(\mathrm{C_6H_6})ΔHcomb∘​=[6ΔHf∘​(CO2​)+3ΔHf∘​(H2​O)]−ΔHf∘​(C6​H6​)
  1. Substitute the given values
ΔHf∘(CO2)=−393.5 kJ mol−1\Delta H_f^\circ(\mathrm{CO_2}) = -393.5\,\text{kJ mol}^{-1}ΔHf∘​(CO2​)=−393.5kJ mol−1 ΔHf∘(H2O(l))=−286 kJ mol−1\Delta H_f^\circ(\mathrm{H_2O(l)}) = -286\,\text{kJ mol}^{-1}ΔHf∘​(H2​O(l))=−286kJ mol−1 ΔHf∘(C6H6(l))=+48.5 kJ mol−1\Delta H_f^\circ(\mathrm{C_6H_6(l)}) = +48.5\,\text{kJ mol}^{-1}ΔHf∘​(C6​H6​(l))=+48.5kJ mol−1

So,

ΔHcomb∘=[6(−393.5)+3(−286)]−48.5\Delta H^\circ_{\text{comb}} = \left[6(-393.5) + 3(-286)\right] - 48.5ΔHcomb∘​=[6(−393.5)+3(−286)]−48.5
  1. Calculate for 1 mole of benzene

First,

6(−393.5)=−23616(-393.5) = -23616(−393.5)=−2361 3(−286)=−8583(-286) = -8583(−286)=−858

Therefore,

ΔHcomb∘=(−2361−858)−48.5\Delta H^\circ_{\text{comb}} = (-2361-858)-48.5ΔHcomb∘​=(−2361−858)−48.5 ΔHcomb∘=−3219−48.5=−3267.5 kJ\Delta H^\circ_{\text{comb}} = -3219 - 48.5 = -3267.5\,\text{kJ}ΔHcomb∘​=−3219−48.5=−3267.5kJ

So for 1 mole benzene,

ΔHcomb∘=−3267.5 kJ\Delta H^\circ_{\text{comb}} = -3267.5\,\text{kJ}ΔHcomb∘​=−3267.5kJ
  1. For 2 moles of benzene
ΔH∘=2×(−3267.5)=−6535 kJ\Delta H^\circ = 2\times (-3267.5) = -6535\,\text{kJ}ΔH∘=2×(−3267.5)=−6535kJ

Thus,

x=6535x = 6535x=6535
  1. Comparison with stored answer

Stored correct answer = 653565356535

Our derived answer also is 653565356535, so they agree.

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