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Thermodynamics question

2023 · 31 Jan · Shift 2 · Q17
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Thermodynamics question

2023 · 31 Jan · Shift 2 · Q17

JEE MainChemistryThermodynamicsNumerical+4 / −1
Enthalpies of formation of CCl4( g),H2O(g),CO2( g)\mathrm{CCl}_{4}(\mathrm{~g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{~g})CCl4​( g),H2​O(g),CO2​( g) and HCl(g)\mathrm{HCl}(\mathrm{g})HCl(g) are −105,−242,−394-105,-242,-394−105,−242,−394 and −92 kJmol−1-92 ~\mathrm{kJ}\mathrm{mol}^{-1}−92 kJmol−1 respectively. The magnitude of enthalpy of the reaction given below is ‾\underline{\hspace{2cm}}​kJ mol−1\mathrm{kJ} ~\mathrm{mol}^{-1}kJ mol−1. (nearest integer) CCl4( g)+2H2O(g)→CO2( g)+4HCl(g)\mathrm{CCl}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{HCl}(\mathrm{g})CCl4​( g)+2H2​O(g)→CO2​( g)+4HCl(g)
Numerical answer
View written solutionFree

Correct answer: 173

  1. Use enthalpy of reaction formula

For the reaction

CCl4(g)+2H2O(g)→CO2(g)+4HCl(g)\mathrm{CCl_4(g)} + 2\mathrm{H_2O(g)} \rightarrow \mathrm{CO_2(g)} + 4\mathrm{HCl(g)}CCl4​(g)+2H2​O(g)→CO2​(g)+4HCl(g)

we use

ΔHrxn=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{\text{rxn}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})ΔHrxn​=∑ΔHf∘​(products)−∑ΔHf∘​(reactants)
  1. Write given enthalpies of formation
ΔHf∘(CCl4(g))=−105 kJ mol−1\Delta H_f^\circ(\mathrm{CCl_4(g)}) = -105\ \text{kJ mol}^{-1}ΔHf∘​(CCl4​(g))=−105 kJ mol−1 ΔHf∘(H2O(g))=−242 kJ mol−1\Delta H_f^\circ(\mathrm{H_2O(g)}) = -242\ \text{kJ mol}^{-1}ΔHf∘​(H2​O(g))=−242 kJ mol−1 ΔHf∘(CO2(g))=−394 kJ mol−1\Delta H_f^\circ(\mathrm{CO_2(g)}) = -394\ \text{kJ mol}^{-1}ΔHf∘​(CO2​(g))=−394 kJ mol−1 ΔHf∘(HCl(g))=−92 kJ mol−1\Delta H_f^\circ(\mathrm{HCl(g)}) = -92\ \text{kJ mol}^{-1}ΔHf∘​(HCl(g))=−92 kJ mol−1
  1. Calculate total enthalpy of products

Products are CO2(g)\mathrm{CO_2(g)}CO2​(g) and 4HCl(g)4\mathrm{HCl(g)}4HCl(g):

∑ΔHf∘(products)=(−394)+4(−92)\sum \Delta H_f^\circ(\text{products}) = (-394) + 4(-92)∑ΔHf∘​(products)=(−394)+4(−92) =−394−368=−762 kJ mol−1= -394 - 368 = -762\ \text{kJ mol}^{-1}=−394−368=−762 kJ mol−1
  1. Calculate total enthalpy of reactants

Reactants are CCl4(g)\mathrm{CCl_4(g)}CCl4​(g) and 2H2O(g)2\mathrm{H_2O(g)}2H2​O(g):

∑ΔHf∘(reactants)=(−105)+2(−242)\sum \Delta H_f^\circ(\text{reactants}) = (-105) + 2(-242)∑ΔHf∘​(reactants)=(−105)+2(−242) =−105−484=−589 kJ mol−1= -105 - 484 = -589\ \text{kJ mol}^{-1}=−105−484=−589 kJ mol−1
  1. Find enthalpy of reaction
ΔHrxn=−762−(−589)\Delta H_{\text{rxn}} = -762 - (-589)ΔHrxn​=−762−(−589) =−762+589=−173 kJ mol−1= -762 + 589 = -173\ \text{kJ mol}^{-1}=−762+589=−173 kJ mol−1
  1. Magnitude of enthalpy

The question asks for the magnitude, so

∣ΔHrxn∣=173 kJ mol−1|\Delta H_{\text{rxn}}| = 173\ \text{kJ mol}^{-1}∣ΔHrxn​∣=173 kJ mol−1

Therefore, the required integer is:

173\boxed{173}173​
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