Mol of Kr gas = 1.244
<h3>Further explanation</h3>In general, the gas equation can be written
<h3> PV=nRT </h3>where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08205 L.atm / mol K
T = temperature, Kelvin
P=1.31 atm
V=23.3 L
T=26+273=299 K
mol of sample :
The question is incomplete, here is the complete question:
A reaction
has a standard free-energy change of -4.88 kJ/mol at 25°C
What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction their concentrations are 0.30 M, 0.40 M and 0 M respectively?
<u>Answer:</u> The equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.
<u>Explanation:</u>
Relation between standard Gibbs free energy and equilibrium constant follows:
where,
= Standard Gibbs free energy = -4.88 kJ/mol = -4880 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant =
T = temperature =
= equilibrium constant of the reaction
Putting values in above equation, we get:
We are given:
Initial concentration of A = 0.30 M
Initial concentration of B = 0.40 M
Initial concentration of C = 0 M
The chemical reaction follows:
<u>Initial:</u> 0.30 0.40 0
<u>At eqllm:</u> 0.30-x 0.40-x x
The expression of equilibrium constant for the above reaction follows:
We are given:
Putting values in above equation, we get:
Neglecting the value of x = 0.657, because change cannot be greater than the initial concentration
So, equilibrium concentration of A =
Equilibrium concentration of B =
Equilibrium concentration of C =
Hence, the equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.