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Thermodynamics — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics5 Marks
Question
  1. Define standard enthalpy of formation. Explain why the enthalpy changes for the reaction given below are not enthalpies of formation of CaCO3 and HBr:
  1. $\text{CaO(s)}+\text{CO}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ \ }\ \text{CaCO}_3(\text{s});$ $\Delta_\text{r}\text{H}^\circ=-178.3\text{kJ mol}^{-1}$
  2. ​​​​​​​$\text{H}_2(\text{g})+\text{Br}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ 2\text{HBr(g)};$ $\Delta_\text{r}\text{H}^\circ=-72.8\text{kJ mol}^{-1}$
  1. Use the bond enthalpies listed below to determine the enthalpy of reaction:

$\ \ \ \ \ \ \ \ \text{H}\\ \ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{H(g)}+2\text{O}=\text{O(g)}\overrightarrow{\ \ \ \ \ }\ \text{O}=\text{C}=\text{O(g)}\\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H}\\+2\text{H}-\text{O}-\text{H(g)}$

Bond enthalpy $(\Delta\text{H}^\circ)/\text{kJ mol}^{-1}$ of C=O = 741; C-H = 414; H-O = 464; O=O = 498.

✓

Answer

  1. Standard enthalpy of formation is defined as enthalpy change when 1 mole of compound is formed from the constituting elements in their standard states.
  1. Is not enthalpy of formation of CaCO3 because it is not being formed from constituting elements.
  2. Is not enthalpy of formation of HBr because 2 moles of HBr are being formed.
  1. $\ \ \ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{H}+2\text{O}=\text{O(g)}\overrightarrow{\ \ \ \ \ \ }\text{O}=\text{C}=\text{O(g)}\\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H}\\+2\text{H}-\text{O}-\text{H(g)}$

$\Delta\text{H}=$ Bond energy of reactants - Bond energy of products

$=2\text{B}_{\text{O}=\text{O}}+4\text{B}_{\text{C}-\text{H}}-2\text{B}_{\text{C}=\text{O}}-4\text{B}_{\text{O}-\text{H}}$

$=2\times498+4\times414-2\times741-4\times464$

$=(996+1656-1482-1856)$

$=-686\text{kJ mol}^{-1}$

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