The reaction, `C(s)+H_2O(g)toCO(g)+H_2(g)` is allowed to occur with 0.1 mol of `H_2O(g)` at `700^(@)`C in a closed vessel of volume 1L at equilibrium. If the partial pressure of `H_2O`(g) is found to decrease by 5.6 atm, then find the value of `K_P` for the reaction. What would be the minimum amount of C( s) so that the above equilibrium can establish?

Explanation

Initial pressure of `H_2O(g)=(nRT)/(V)=(0.1xx0.0821xx973)/1`
=7.98 atm
Partial pressure `C(s)+H_2O(g) At equilibrium: `7.98-5.6" "5.6atm" "5.6atm`
=2.38atm
`K_p=(p_(CO)xxp_(H_2))/(p_(H_2O))=(5.6xx5.6)/(2.38)atm=13.18atm`
Number of moles of `CO_2` at equilibrium:
`=(p_(CO_2)V)/(RT)=(5.6xx1)/(0.0821xx973)=0.7mol`
`0.07 mol CO_2-=0.07molC(s)-=0.07xx12g-=0.84g`