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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
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B
2
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C
3
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D
4
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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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