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Question

The degree of dissociation of $$N_{2}O_{4}$$ into $$NO_{2}$$ at $$1.5$$ atmosphere and $$40^{\circ}C$$ is $$0.25$$. Calculate its $$K_{P}$$ at $$40^{\circ}C$$. Also report degree of dissociation at $$10$$ atmospheric pressure at same temperature.


Solution

$$\quad \quad \quad \quad \quad \quad \quad  { N }_{ 2 }{ O }_{ 4 }\leftrightharpoons 2N{ O }_{ 2 }\\ initially\quad \quad \quad \quad 1\quad \quad \quad  \quad 0\\ at\quad equlibrium\quad 1-\alpha \quad \quad  2\alpha $$

Total moles of gas = 1+$$\alpha$$

If P is the total pressure, then partial pressure are :-

From Roult's law we have,

Partial pressure of the gas-

$$p(g)=P_T\times x$$
$${ P }_{ { N }_{ 2 }{ O }_{ 4 } }=\cfrac { P(1-\alpha ) }{ (1+\alpha ) } \\ { P }_{ N{ O }_{ 2 } }=\cfrac { P(2\alpha ) }{ (1+\alpha ) } \\ { K }_{ P }=\cfrac { { ({ P }_{ N{ O }_{ 2 } }) }^{ 2 } }{ { P }_{ { N }_{ 2 }{ O }_{ 4 } } } =\cfrac { { \left[ \cfrac { P(2\alpha ) }{ (1+\alpha ) }  \right]  }^{ 2 } }{ \left[ \cfrac { P(1-\alpha ) }{ (1+\alpha ) }  \right]  } =\cfrac { { (P(2\alpha )) }^{ 2 } }{ P(1-{\alpha}^2 ) } =\cfrac { 1atm{ (2\times (0.25)) }^{ 2 } }{ 1-{0.25}^2 } =\cfrac { 1atm\times (0.25) }{ 0.9375 } =0.26\quad atm$$

$$\therefore K_{P}$$ at 40℃ is 0.26 atm.  

Chemistry

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