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Thermodynamics question

2021 · 17 Mar · Shift 1 · Q21
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Thermodynamics question

2021 · 17 Mar · Shift 1 · Q21

JEE MainChemistryThermodynamicsNumerical+4 / −1
The standard enthalpies of formation of Al2O3Al_2O_3Al2​O3​ and CaOCaOCaO are −-− 1675 kJ mol-1 and −-− 635 kJ mol −-− 1 respectively. For the reaction 3CaOCaOCaO + 2AlAlAl →\to→ 3CaCaCa + Al2O3Al_2O_3Al2​O3​ the standard reaction enthalpy Δ\DeltaΔ rH0 = ‾\underline{\hspace{2cm}}​ kJ. (Round off to the Nearest Integer)
Numerical answer
View written solutionFree

Correct answer: 230

  1. Write the reaction

3CaO+2Al→3Ca+Al2O33CaO + 2Al \rightarrow 3Ca + Al_2O_33CaO+2Al→3Ca+Al2​O3​

We use

ΔrH∘=∑ν ΔfH∘(products)−∑ν ΔfH∘(reactants)\Delta_r H^\circ = \sum \nu \, \Delta_f H^\circ(\text{products}) - \sum \nu \, \Delta_f H^\circ(\text{reactants})Δr​H∘=∑νΔf​H∘(products)−∑νΔf​H∘(reactants)

  1. Given standard enthalpies of formation

ΔfH∘(Al2O3)=−1675 kJ mol−1\Delta_f H^\circ(Al_2O_3) = -1675\ \text{kJ mol}^{-1}Δf​H∘(Al2​O3​)=−1675 kJ mol−1 ΔfH∘(CaO)=−635 kJ mol−1\Delta_f H^\circ(CaO) = -635\ \text{kJ mol}^{-1}Δf​H∘(CaO)=−635 kJ mol−1

For elements in their standard states:

ΔfH∘(Al)=0,ΔfH∘(Ca)=0\Delta_f H^\circ(Al) = 0, \qquad \Delta_f H^\circ(Ca) = 0Δf​H∘(Al)=0,Δf​H∘(Ca)=0

  1. Calculate sum for products

Products are 3Ca3Ca3Ca and 1Al2O31Al_2O_31Al2​O3​:

∑ΔfH∘(products)=3(0)+1(−1675)=−1675 kJ\sum \Delta_f H^\circ(\text{products}) = 3(0) + 1(-1675) = -1675\ \text{kJ}∑Δf​H∘(products)=3(0)+1(−1675)=−1675 kJ

  1. Calculate sum for reactants

Reactants are 3CaO3CaO3CaO and 2Al2Al2Al:

∑ΔfH∘(reactants)=3(−635)+2(0)=−1905 kJ\sum \Delta_f H^\circ(\text{reactants}) = 3(-635) + 2(0) = -1905\ \text{kJ}∑Δf​H∘(reactants)=3(−635)+2(0)=−1905 kJ

  1. Find reaction enthalpy

ΔrH∘=(−1675)−(−1905)\Delta_r H^\circ = (-1675) - (-1905)Δr​H∘=(−1675)−(−1905)

ΔrH∘=−1675+1905=230 kJ\Delta_r H^\circ = -1675 + 1905 = 230\ \text{kJ}Δr​H∘=−1675+1905=230 kJ

  1. Final answer

230\boxed{230}230​

So, the standard reaction enthalpy is 230230230 kJ.

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