A vessel at 1000 K contains `CO_(2)` with a pressure of 0.5 atm. Some of the `CO_(2)` is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :

Options

A. 1.8 atm
B. 3 atm
C. 0.3 atm
D. 0.18 atm

Answer

A

Explanation

`{:(,CO_(2),+,C_("grapnite"),hArr,2CO),(P_("intial"),0.5atm,,,,0),(p_("final"),0.5-xatm,,,,2x atm):}`
Total P at equilibrium `=0.5-x+2x=0.5+x atm`
`therefore x=0.8-0.5=0.3 atm`
Now `k_(p)=(P_(CO))^(2)//P_(CO_(2))`
`=((2xx0.3)^(2))/((0.5-0.3))=((0.6)^(2))/((0.2))=1.8 atm`