Question

$$1$$mole of $$NO$$ and $$1$$ mole of $$O_3$$ are taken in $$10L$$ vessel and heated. At equilibrium, $$50\%$$ of $$NO$$ (by mass) reacts with $$O_3$$ according to the equation:$$NO(g) + O_{3(g)} \rightleftharpoons NO_{2(g)} + O_{2(g)}$$.What will be the equilibrium constant for this reaction?

A
1
B
2
C
3
D
4

Solution

The correct option is A $$1$$The given reaction is :-                                  $$NO\left( g \right) +{ O }_{ 3 }\left( g \right) \rightleftharpoons { NO }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right)$$Initial cone :                  $$1$$               $$1$$                 $$0$$                 $$0$$At eqm :                $$1-x$$         $$1-x$$                $$x$$                 $$x$$Given initial conc. of $$NO\quad \& \quad { O }_{ 3 }$$ each have reacted.Now,   $$x=50$$% of $$NO=\dfrac { 50 }{ 100 } \times 1=\dfrac { 1 }{ 2 }$$So,  $${ K }_{ C }=\dfrac { \left[ { NO }_{ 2 } \right] \left[ { O }_{ 2 } \right] }{ \left[ NO \right] \left[ { O }_{ 3 } \right] }$$$${ K }_{ C }=\dfrac { 1/2\times 1/2 }{ 1/2\times 1/2 } =1$$Chemistry

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