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QJEEM11Ch49ac: JEE Main 2011 Chemistry - Class 11 JEEM11Ch49ac Solution: A vessel at 1000 K contains CO 2 with a pressure of 0.5 atm.…
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :
Multiple Choice Options
A
3 atm
B
0.3 atm
C
0.18 atm
D
1.8 atm
Step-by-Step Verified Solution
C O 2 + C ( g r a p n i t e ) ⇌ 2 C O C{O_2} + {C_{\left( {grapnite} \right)}}\,\rightleftharpoons\,2CO C O 2 + C ( g r a p ni t e ) ⇌ 2 CO P i n i t i a l 0.5 a t m 0 {P_{initial}}\,\,0.5atm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 P ini t ia l 0.5 a t m 0 P f i n a l ( 0.5 − x ) a t m 2 x a t m {P_{final}}\,\,\,\left( {0.5 - x} \right)atm\,\,\,\,\,\,\,2x\,atm P f ina l ( 0.5 − x ) a t m 2 x a t m Total
P P P at equilibrium
= 0.5 − x + 2 x = 0.5 + x a t m = 0.5 - x + 2x = 0.5 + x\,atm = 0.5 − x + 2 x = 0.5 + x a t m 0.8 = 0.5 + x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.8 = 0.5 + x 0.8 = 0.5 + x ∴ \therefore ∴ x = 0.8 − 0.5 = 0.3 a t m \,\,\,x = 0.8 - 0.5 = 0.3\,atm x = 0.8 − 0.5 = 0.3 a t m Now
k p = ( P C O ) 2 / P C O 2 \,\,\,{k_p} = {\left( {{P_{CO}}} \right)^2}/{P_{C{O_2}}} k p = ( P CO ) 2 / P C O 2 = ( 2 × 0.3 ) 2 ( 0.5 − 0.3 ) = ( 0.6 ) 2 ( 0.2 ) = {{{{\left( {2 \times 0.3} \right)}^2}} \over {\left( {0.5 - 0.3} \right)}} = {{{{\left( {0.6} \right)}^2}} \over {\left( {0.2} \right)}} = ( 0.5 − 0.3 ) ( 2 × 0.3 ) 2 = ( 0.2 ) ( 0.6 ) 2 = 1.8 a t m = 1.8\,atm = 1.8 a t m
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