What is the enthalpy of the overall chemical reaction ch4(g)+4cl2(g)- ccl4(g)+4hcl(g)?
1 answer:
Answer:
-205.7kj
Explanation:
Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of the reaction.
The expression for enthalpy for the following reaction will be,
where,
n = number of moles
Now put all the given values in the above expression, we get:
Therefore, the enthalpy of the following reaction is, -205.7kj
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Answer:
The value of is 0.02495.
Explanation:
Initial concentration of gas = 0.675 M
Initial concentration of gas = 0.973 M
Equilibrium concentration of mustard gas = 0.35 M
initially
0.675 M 0.973 M 0
At equilibrium ;
(0.675-0.35) M (0.973-2 × 0.35) M 0.35 M
The equilibrium constant is given as :
The relation between and
are :
where,
= equilibrium constant at constant pressure = ?
= equilibrium concentration constant =14.45
R = gas constant = 0.0821 L⋅atm/(K⋅mol)
T = temperature = 20.0°C =20.0 +273.15 K=293.15 K
= change in the number of moles of gas = [(1) - (1 + 2)]=-2
Now put all the given values in the above relation, we get:
The value of is 0.02495.