The ideal gas law only approximates the behaviour of real gases. The conditions
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Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 -
Solving
-=75.5 kJ/mol - 2*90.25 kJ/mol
-=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>
The standard ambient temperature and pressure are
Temperature =298 K
Pressure = 1atm
The density of gas is 1.5328 g/L
density = mass of gas per unit volume
the ideal gas equation is
PV = nRT
P = pressure = 1 atm
V = volume
n = moles
R= gas constant = 0.0821 Latm/mol K
T = 298 K
moles = mass / molar mass
so we can write
n/V = density / molar mass
Putting values
Thus molar mass of gas is 37.50g/mol