The reaction system
1 answer:
Answer:
e. As equilibrium is reestablished, the partial pressure of hydrogen decreases.
Explanation:
Hello,
In this case, for the given chemical reaction:
Considering the Le Chatelier's principle which states that adding more reactant shifts the reaction towards the products, condition that is related with the doubling of the hydrogen partial pressure, we can notice that e. As equilibrium is reestablished, the partial pressure of hydrogen decreases since the addition of hydrogen will be used to form more methane and hydrogen sulfide, for that reason, its amount is decreased to reattain equilibrium.
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The question is incomplete, the table of the question is given below
Answer:
I) xA= 0.34, yA= 0.55
ii) 76.2 mole % vapor
iii) Percentage of vapor volume = 98%
Explanation:
i) xA= 0.34, yA= 0.55
xA= 0.34, yA= 0.55
ii) 0.50 = 0.55 nv + 0.34 nL
Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid
This shows 76.2 mole % vapor
iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3
Therefore, ρ = 0.790 g/cm3
Now, we have:
MA = 58.08 g/mol and ME= 46.07 g/mol
So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol
1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor
Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3
Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3
Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %
