Question

Graphite can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 KK is Kp=1.60×10−3

If a 1.55-LL reaction vessel initially contains 157 mbar of water at 700.0 KK in contact with excess graphite, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium.

Answer 1

Answer:

Answer:- => pressure = 249 torr = 249/760 = 0.3276 atm => C + H2O ---> CO + H2 => initial pressure of H2O = 0.3276 atm => let equilibrium pressure of CO = X => equalibrium pressure of…

Answer 2

0.3476% of mass of hydrogen gas of the gaseous reaction mixture at equilibrium.

Explanation:

• The equilibrium constant for the reaction at 700.0 KK is Kp=$1.60$×$10^{-3}$
• Graphite can react with gaseous water to form carbon monoxide gas and hydrogen gas.
• If a $1.55-L$ reaction vessel initially contains $157 mbar$ of water at $700.0 KK$
• given reaction is
• $C + H2O ---> CO + H2$
• pressure = 249 torr = 249/760 = 0.3276
• atm initial pressure of H2O = 0.3276 atm
• 0.3476% of mass of hydrogen gas of the gaseous reaction mixture at equilibrium.