1.

At 1090K, K_(p) for the reaction CO_(2)(g) +C(s) rarr 2CO(g) is 10 atm, At the constant temperature ,Equilibrium position will shift, when pressure change occurs. In the above Equilibrium reaction, partial pressure of CO_(2) of equilibrium :

Answer»

10 atm
0.938 atm
0.23 atm
0.77 atm

Solution :`CO_(2)(g) + C(s) rarr 2CO(g)`
102Initial moles
1-x02xmoles at EQUILIBRIUM
`k_(p) = ((p_(CO))^(2))/( P_(CO_(2)))= (((2x)/( 1+x).P)^(2))/((1-x)/(1+x).P) = 10 rarr x = 0.62 `
Moles of `CO_(2) = 1-x = 1 - 0.62 = 0.38`
`P_(CO_(2)) = ( 1-x)/( 1+x ) . P = ( 1- 0.62 )/( 1+ 0.62) ( 4) = 0.928` atm


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