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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Sample Paper Topic Test - Thermo Chemistry (Thermodynamics)

  • question_answer
    Calculate the heat change in the reaction \[4\,N{{H}_{3}}(g)+3{{O}_{2}}(g)\to 2{{N}_{2}}(g)+6{{H}_{2}}O(\ell )\] at 298 K, given that the heats of formation at 298 K  for \[N{{H}_{3}}(g)\] and \[{{H}_{2}}O(\ell )\] are - 46.0 and - 28.60 kJ \[mo{{l}^{-1}}\] respectively.

    A) - 1861 kJ

    B) - 1361 kJ

    C) - 1261 kJ

    D) - 1532 kJ

    Correct Answer: D

    Solution :

    [d] \[\Delta {{H}^{o}}\] for the reaction
    \[4N{{H}_{3}}\left( g \right)+3{{O}_{2}}\left( g \right)\to 6{{H}_{2}}O\left( l \right)+2{{N}_{2}}\left( g \right)\]
    \[\Delta {{H}^{o}}=\Delta H_{f}^{o}\left( \text{products} \right)-\Delta H_{f}^{o}\left( \text{reactants} \right)\]
    \[=\left\{ 6H_{f}^{o}\left[ {{H}_{2}}O\left( l \right) \right]+\Delta {{H}^{\circ }}_{f}\left[ {{N}_{2}}\left( g \right) \right] \right\}\]
    \[-\left\{ 4\Delta H\left[ N{{H}_{3}}\left( g \right) \right]+3\Delta {{H}_{f}}\left[ {{O}_{2}}\left( g \right) \right] \right\}\]
    \[\Delta H_{f}^{o}\left[ {{H}_{2}}O\left( l \right) \right]=-286.0\,\,kJ\,\,mo{{l}^{-1}}\]
    \[\Delta H_{f}^{o}\left[ N{{H}_{2}}\left( g \right) \right]=0\] and \[\Delta H_{f}^{o}\left[ {{N}_{2}}\left( g \right) \right]\]
    = 0 (by convention)
    \[\Delta {{H}^{o}}=\left\{ 6\left( -286 \right)+2\left( 0 \right) \right\}-\left\{ 4\left( -46.0 \right)+3\left( 0 \right) \right\}\]
    \[=-1716+184=-1532\,kJ\]


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