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Thermodynamics question

2022 · 24 Jun · Shift 2 · Q4
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  5. /2022 · 24 Jun · Shift 2 · Q4

Thermodynamics question

2022 · 24 Jun · Shift 2 · Q4

JEE MainChemistryThermodynamicsMCQ+4 / −1

At 25 ∘^\circ∘ C and 1 atm pressure, the enthalpies of combustion are as given below :

Substance H2{H_2}H2​ C (graphite) C2H6(g){C_2}{H_6}(g)C2​H6​(g)
ΔcHΘkJ mol−1{{{\Delta _c}{H^\Theta }} \over {kJ\,mo{l^{ - 1}}}}kJmol−1Δc​HΘ​ −286.0- 286.0−286.0 −394.0- 394.0−394.0 −1560.0- 1560.0−1560.0

The enthalpy of formation of ethane is

  1. A
    +54.0 kJ mol −-− 1
  2. B
    −-− 68.0 kJ mol −-− 1
  3. C
    −-− 86.0 kJ mol −-− 1
  4. D
    +97.0 kJ mol −-− 1
View written solutionFree

Correct answer: C

  1. Write the formation reaction of ethane

The standard enthalpy of formation of ethane corresponds to:

2 C(graphite)+3 H2(g)→C2H6(g)2\,C(\text{graphite}) + 3\,H_2(g) \rightarrow C_2H_6(g)2C(graphite)+3H2​(g)→C2​H6​(g)

We need to find ΔfH∘(C2H6)\Delta_f H^\circ\big(C_2H_6\big)Δf​H∘(C2​H6​).

  1. Write the combustion reactions and their enthalpies

Given:

  • For hydrogen: H2(g)+12O2(g)→H2O(l),ΔcH∘=−286.0 kJ mol−1H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \qquad \Delta_c H^\circ = -286.0\ \text{kJ mol}^{-1}H2​(g)+21​O2​(g)→H2​O(l),Δc​H∘=−286.0 kJ mol−1

  • For graphite: C(graphite)+O2(g)→CO2(g),ΔcH∘=−394.0 kJ mol−1C(\text{graphite}) + O_2(g) \rightarrow CO_2(g), \qquad \Delta_c H^\circ = -394.0\ \text{kJ mol}^{-1}C(graphite)+O2​(g)→CO2​(g),Δc​H∘=−394.0 kJ mol−1

  • For ethane: C2H6(g)+72O2(g)→2CO2(g)+3H2O(l),ΔcH∘=−1560.0 kJ mol−1C_2H_6(g) + \frac{7}{2}O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l), \qquad \Delta_c H^\circ = -1560.0\ \text{kJ mol}^{-1}C2​H6​(g)+27​O2​(g)→2CO2​(g)+3H2​O(l),Δc​H∘=−1560.0 kJ mol−1

  1. Use Hess's law

If we combust the elements forming ethane:

2C+3H2→2CO2+3H2O2C + 3H_2 \rightarrow 2CO_2 + 3H_2O2C+3H2​→2CO2​+3H2​O

Its enthalpy change is:

2(−394.0)+3(−286.0)=−788.0−858.0=−1646.0 kJ mol−12(-394.0) + 3(-286.0) = -788.0 - 858.0 = -1646.0\ \text{kJ mol}^{-1}2(−394.0)+3(−286.0)=−788.0−858.0=−1646.0 kJ mol−1

Now this overall process can also be written in two steps:

  • Formation of ethane: 2C+3H2→C2H6ΔfH∘=?2C + 3H_2 \rightarrow C_2H_6 \qquad \Delta_f H^\circ = ?2C+3H2​→C2​H6​Δf​H∘=?

  • Combustion of ethane: C2H6+72O2→2CO2+3H2OΔcH∘=−1560.0C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O \qquad \Delta_c H^\circ = -1560.0C2​H6​+27​O2​→2CO2​+3H2​OΔc​H∘=−1560.0

So,

ΔfH∘+(−1560.0)=−1646.0\Delta_f H^\circ + (-1560.0) = -1646.0Δf​H∘+(−1560.0)=−1646.0

Therefore,

ΔfH∘=−1646.0+1560.0=−86.0 kJ mol−1\Delta_f H^\circ = -1646.0 + 1560.0 = -86.0\ \text{kJ mol}^{-1}Δf​H∘=−1646.0+1560.0=−86.0 kJ mol−1

  1. Match with options

ΔfH∘(C2H6)=−86.0 kJ mol−1\boxed{\Delta_f H^\circ\big(C_2H_6\big) = -86.0\ \text{kJ mol}^{-1}}Δf​H∘(C2​H6​)=−86.0 kJ mol−1​

So the correct option is:

C\boxed{\text{C}}C​

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