KVPY SA Exam 2016 Question Paper

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Question : 73 of 80

Marks: +1, -0

Given

\table NO(g) + O_3 (g) →NO_2 (g) + O_2 (g) ,\∆H = –198.9 \kJ\/\mol;O_3(g) →3/2O_2( g), \∆H = –142.3 \kJ\/\mol;O_2 (g) →O(g), \∆H = +495.0 \kJ\/\mol

The entalpy change (∆H) for the following reaction is
NO (g) + O (g) →

\NO_2

(g)

–304.1 kJ/mol
+304.1 kJ/mol
-403.1 kJ/mol
+403.1 kJ/mol

Solution:

\table NO(g) + O_3 (g) →NO_2 (g) + O_2 (g) ,\∆H = –198.9 \kJ\/\mol;O_3(g) →3/2O_2( g), \∆H = –142.3 \kJ\/\mol;O_2 (g) →O(g), \∆H = +495.0 \kJ\/\mol

\Δ\H\of \NO+\O(\g)→\NO_2(g)

For this

\i - \ii - \iii/2

=-198.9-(-142.3)-495/2

= -304.1 \KJ \/ \mole

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