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At 1127 K and 1 atm pressure, a gaseous ...

At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2)` in equilibrium with solid carbon has `90.55% CO` by mass:
`C_((s))+CO_(2(g))hArr2CO_((g))`
Calculate `K_(c)` for the reaction at the above temperature.

A

1.53

B

0.153

C

0.53

D

0.76

Text Solution

Verified by Experts

The correct Answer is:
B
Let the total mass of the mixture of CO and `CO_(2)` is 100 g, then CO=90.55g and `CO_(2)=100-90.55=9.45g`
Moles of `CO=(90.55)/(28)=3.234,`
Moles of `CO_(2)=(9.45)/(44)=0.215`
Mole fraction of `CO=(3.234)/(3.234+0.215)=0.938`
Mole fraction of `CO_(2)=(0.215)/(3.234+0.215)=0.062`
`:." "p` co = mole fraction `xx` total pressure `=0.938xx1` atm `=0.962` atm
`p_(co_(2))=0.062xx1` atm `=0.062` atm
`K_(s)=(p_(co)^(2))/(p_(co_(2)))=((0.938)^(2))/(0.062)=14.19`
Now, `Deltan_(g)=2-1=1,K_(p)=K_(c)(RT)^(Deltan_(g))`
`orK_(c)=(K_(p))/(RT)=(14.19)/(0.0821xx1127)=0.153`
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