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At 1090K, K(p) for the reaction CO(2)(...

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 `:`

A

10 atm

B

0.938 atm

C

0.23 atm

D

0.77 atm

Text Solution

Verified by Experts

The correct Answer is:
B
`CO_(2)(g) + C(s) rarr 2CO(g)`
1 0 2 Initial moles
1-x 0 2x moles 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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Knowledge Check

  • 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 reaction, if we use catalyst. Equilibrium constant value :

    A
    increases
    B
    decreases
    C
    does no change
    D
    change is rapid

  • 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 reaction, at what total pressure will the gaes analyse 8% CO_(2) by volume ?

    A
    10 atm
    B
    0.23 atm
    C
    0.95 atm
    D
    0.45 atm

  • The value of K_(p) for the reaction CO_(2)(g)+C(s)hArr2CO(g) is 3.0 bar at 1000 K . If initially P_(CO_(2)) = 0.48 bar, P_(CO) = 0 bar and pure graphite is present then determine equilibrium partial pressue of CO and CO_(2) .

    A
    `P_(CO)=0.15bar, P_(CO_(2))= 0.66 bar`
    B
    `P_(CO)=0.66 bar, P_(CO_(2))= 0.15 bar`
    C
    `P_(CO)=0.33 bar, P_(CO_(2))= 0.66 bar`
    D
    `P_(CO)= 0.66 bar, P_(CO_(2)) = 0.33 bar`

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