JEE Main 2014 — Chemistry Physical Chemistry

In case of heterogeneous equilibrium. The achive mass of solid is unit so not consider in the expression of ${\text{K}}_{\text{p}} \text{and} {\text{K}}_{\text{c}}$Kp​andKc​

$${\text{NH}}_{2} {\text{COONH}}_{4} ( \text{s} ) \rightleftharpoons 2 {\text{NH}}_{3} ( \text{g} ) + {\text{CO}}_{2} ( \text{g} )$$NH2​COONH4​(s)⇌2NH3​(g)+CO2​(g)

At equilibrium $\Rightarrow$⇒ 2 p and p will be partial pressures of ammonia and carbon dioxide gas respectively.

$${\text{K}}_{\text{P}} \text{=} { [ {\text{P}}_{ { \text{NH}}_{\text{3}} } ]}^{\text{2}} [ {\text{P}}_{ { \text{CO}}_{\text{2}} } ] \text{=} { [ \text{2P}]}^{\text{2}} [\text{P}] \text{=} { \text{4P}}^{\text{3}}$$KP​=[PNH3​​]2[PCO2​​]=[2P]2[P]=4P3

$$4 {\text{P}}^{3} = 2 \cdot 9 \times 1 {0}^{ - 5 } {\text{atm}}^{3}$$4P3=2⋅9×10−5atm3

$${\text{P}}^{3} = 0 \cdot 7 2 5 \times 1 {0}^{ - 5 } {\text{atm}}^{3} = 7 \cdot 2 5 \times 1 {0}^{ - 6 } {\text{atm}}^{3}$$P3=0⋅725×10−5atm3=7⋅25×10−6atm3

$$\text{P} = \sqrt[ 3 ]{ 7 \cdot 2 5 \times 1 {0}^{ - 6 } } = 1 \cdot 9 4 \times 1 {0}^{ - 2 } \text{atm}$$P=37⋅25×10−6​=1⋅94×10−2atm

$$\Sigma \text{P} = { ( {\text{P}}_{\text{Total}} ) }_{\text{equilibrium}} = 2 \text{P} + \text{P} = 3 \text{P}$$ΣP=(PTotal​)equilibrium​=2P+P=3P

=$3 \times 1 \cdot 9 4 \times 1 {0}^{ - 2 } \text{atm}$3×1⋅94×10−2atm

=$5 \cdot 8 2 \times 1 {0}^{ - 2 } \text{atm}$5⋅82×10−2atm