The equilibrium constant is 0.0022.
Explanation:
The values given in the problem is
ΔG° = 1.22 ×10⁵ J/mol
T = 2400 K.
R = 8.314 J mol⁻¹ K⁻¹
The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.
The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K
So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.
We get,
So, the equilibrium constant is 0.0022.
<h3>
Answer:</h3>
382.63 K
<h3>
Explanation:</h3>
We are given;
- Volume of Iodine as 71.4 mL
- Mass of Iodine as 0.276 g
- Pressure of Iodine as 0.478 atm
We are required to calculate the temperature of Iodine
- We are going to use the ideal gas equation;
- According to the ideal gas equation; PV = nRT, where R is the ideal gas constant, 0.082057 L.atm/mol.K.
T = PV ÷ nR
But, n, the number of moles = Mass ÷ Molar mass
Molar mass of iodine = 253.8089 g/mol
Thus, n = 0.276 g ÷ 253.8089 g/mol
= 0.001087 moles
Therefore;
T = (0.478 atm × 0.0714 L) ÷ (0.001087 moles × 0.082057)
= 382.63 K
Thus, the temperature of Iodine in Kelvin is 382.63 K
a) 1 mole of Ne
b) i/2 mole of Mg
c) 1570 moles of Pb.
d) 2.18125*10^-13 moles of oxygen.
Explanation:
The number of moles calculated by Avogadro's number in 6.23*10^23 of Neon.
6.23*10^23= 1/ 6.23*10^23
= 1 mole
The number of moles calculated by Avogadro's number in 3.01*10^23 of Mg
3.2*10^23=1/6.23*10^23
= 1/2 moles of Pb.
Number of moles in 3.25*10^5 gm of lead.
atomic weight of Pb=
n=weight/atomic weight
= 3.25*10^5/ 207
= 1570 moles of Pb.
Number of moles 4.50 x 10-12 g O
number of moles= 4.50*10^-12/16
= 2.18125*10^-13 moles of oxygen.
Density = mass / volume so
Density = 156 / 20.0
Density = 7.8 g/mL
