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?

The reaction, C(s)+H_2O(g)toCO(g)+H_2(g) is allowed to occur with 0.1

1.	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?
Answer» SOLUTION :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`