Methane gas and steam in equimolar amounts are taken in a flask and sealed where the ollowing equilibrium was established: IODcoe abw H(g)+HO(v)Co(g)+34 (9)2t (g)-Co(g)CH,H. on
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Answer : The equilibrium constant for the reaction is 0.75
Explanation :
The chemical equation for the reaction can be written as follows.
We are taking equimolar amounts of methane and steam. So let's assume we have x moles of methane and steam initially.
Let us set up the ICE table for this reaction.
From the ICE table, we can say that
Equilibrium partial pressure of CO is y. But we have given that equilibrium partial pressure of CO is 0.5 atm.
Therefore we have y = 0.5 atm
From ICE table we have,
total pressure =
But the total pressure is given as 5 atm.
Therefore ,
But y = 0.5, therefore we have
Partial pressures of the gases at equilibrium are as follows.
P[CH₄] = x-y = 2.0 - 0.5 = 1.5 atm
P[H₂O] = x-y = 2.0 - 0.5 = 1.5 atm
P[CO] = y = 0.5 atm
P[H₂} = 3y = 3(0.5) = 1.5 atm
The equilibrium constant is calculated as,
Let us plug in the above equilibrium values.
The equilibrium constant for the reaction is 0.75
Explanation :
The chemical equation for the reaction can be written as follows.
We are taking equimolar amounts of methane and steam. So let's assume we have x moles of methane and steam initially.
Let us set up the ICE table for this reaction.
From the ICE table, we can say that
Equilibrium partial pressure of CO is y. But we have given that equilibrium partial pressure of CO is 0.5 atm.
Therefore we have y = 0.5 atm
From ICE table we have,
total pressure =
But the total pressure is given as 5 atm.
Therefore ,
But y = 0.5, therefore we have
Partial pressures of the gases at equilibrium are as follows.
P[CH₄] = x-y = 2.0 - 0.5 = 1.5 atm
P[H₂O] = x-y = 2.0 - 0.5 = 1.5 atm
P[CO] = y = 0.5 atm
P[H₂} = 3y = 3(0.5) = 1.5 atm
The equilibrium constant is calculated as,
Let us plug in the above equilibrium values.
The equilibrium constant for the reaction is 0.75
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