A sample of water gas has a composition by volume of 50% H2, 45% CO and 5% CO2. Calculate the volume in litres at STP at which the water gas on treatment with excess of steam will produce 5 L of H2. The equation for the reaction is: CO(g)+H2O(g)→CO2(g)+H2(g)
A sample of water gas has a composition by volume of 50% H2, 45% CO and 5% CO2. Calculate the volume in litres at STP at which the water gas on treatment with excess of steam will produce 5 L of H2. The equation for the reaction is: CO(g)+H2O(g)→CO2(g)+H2(g)
The correct option is A 5.26 L
Given information:
Composition of the water gas is 50% H2, 45% CO and 5% CO2
Let the volume of water gas be x L. Then,
volume of H2 = 0.5x L
volume of CO =0.45x L
volume of CO2 = 0.05x L
Final volume of H2 after reaction with steam = 5 L
CO in water gas reacts with steam to produce H2 according to the equation,
CO+H2O→CO2+H2
Applying Gay Lussac's law,
volume of H2 formed = volume of CO consumed = 0.45x L
Final volume of H2 =
(volume of H2 present initially + volume H2 formed from CO)
⇒5 L=0.5x L+0.45x L
⇒x=5.26 L
Theory :
Gay-Lussac’s Law: At constant V and n,
P∝T
P=k3T
P1P2=T1T2
Isochor: Constant volume plot
Case 1: P vs T(K)
From Gay-Lussac’s law :
P=k3T
It is similar to the equation of the straight line passing through the origin.
y=mx
Where y=P , m=k3 and x=T
Case 2: P vs T(K) at different volumes
Since by increasing the pressure, volume increases so for the isochore V1 pressure is maximum so volume is least and it gradually increases with decrease in pressure and volume is maximum in isochore V4.
The correct order of volume :
V1<V2<V3<V4
Case 3: P vs t(∘C)
From Gay lussac’s law :
P=k3T
Put T=(t+273.15)K
P=k3(273.15)+k3t
It is similar to equation of straight line y=mx+c
Where y=P , intercept(c)=k3(273.15), x=t and m=k3
Case 4: PT vs P
From Gay-Lussac law,
P=k3T
PT=k3
Therefore, on increasing P, PT will remain constant.
Application of Gay-Lussac’s Law:
-
A pressure cooker helps in cooking food faster.
-
It is recommended to fill low air in the tire tubes in summer because as the temperature is increased, the pressure increases. Therefore, the tubes can blow.
-
Aerosol bottles are kept away from heat because as the temperature increases, the pressure increases and it can result in an explosion.
Given information:
Composition of the water gas is 50% H2, 45% CO and 5% CO2
Let the volume of water gas be x L. Then,
volume of H2 = 0.5x L
volume of CO =0.45x L
volume of CO2 = 0.05x L
Final volume of H2 after reaction with steam = 5 L
CO in water gas reacts with steam to produce H2 according to the equation,
CO+H2O→CO2+H2
Applying Gay Lussac's law,
volume of H2 formed = volume of CO consumed = 0.45x L
Final volume of H2 =
(volume of H2 present initially + volume H2 formed from CO)
⇒5 L=0.5x L+0.45x L
⇒x=5.26 L
Theory :
Gay-Lussac’s Law: At constant V and n,
P∝T
P=k3T
P1P2=T1T2
Isochor: Constant volume plot
Case 1: P vs T(K)
From Gay-Lussac’s law :
P=k3T
It is similar to the equation of the straight line passing through the origin.
y=mx
Where y=P , m=k3 and x=T
Case 2: P vs T(K) at different volumes
Since by increasing the pressure, volume increases so for the isochore V1 pressure is maximum so volume is least and it gradually increases with decrease in pressure and volume is maximum in isochore V4.
The correct order of volume :
V1<V2<V3<V4
Case 3: P vs t(∘C)
From Gay lussac’s law :
P=k3T
Put T=(t+273.15)K
P=k3(273.15)+k3t
It is similar to equation of straight line y=mx+c
Where y=P , intercept(c)=k3(273.15), x=t and m=k3
Case 4: PT vs P
From Gay-Lussac law,
P=k3T
PT=k3
Therefore, on increasing P, PT will remain constant.
Application of Gay-Lussac’s Law:
-
A pressure cooker helps in cooking food faster.
-
It is recommended to fill low air in the tire tubes in summer because as the temperature is increased, the pressure increases. Therefore, the tubes can blow.
-
Aerosol bottles are kept away from heat because as the temperature increases, the pressure increases and it can result in an explosion.
