Answer:
0.169
Explanation:
Let's consider the following reaction.
A(g) + 2B(g) ⇄ C(g) + D(g)
We can find the pressures at equilibrium using an ICE chart.
A(g) + 2 B(g) ⇄ C(g) + D(g)
I 1.00 1.00 0 0
C -x -2x +x +x
E 1.00-x 1.00-2x x x
The pressure at equilibrium of C is 0.211 atm, so x = 0.211.
The pressures at equilibrium are:
pA = 1.00-x = 1.00-0.211 = 0.789 atm
pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm
pC = x = 0.211 atm
pD = x = 0.211 atm
The pressure equilibrium constant (Kp) is:
Kp = pC × pD / pA × pB²
Kp = 0.211 × 0.211 / 0.789 × 0.578²
Kp = 0.169
X=0.031903 I think if you don’t know how to do this photo math would be a good thing for you
Answer:
ΔG = 16.218 KJ/mol
Explanation:
- dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate
∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol
∴ R = 8.314 J/K.mol
∴ T = 298 K
∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]
⇒ Q = 0.00300 / 0.100 = 0.03
⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))
⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )
Answer : The mole fraction and partial pressure of 
and 
gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.
Explanation : Given,
Moles of 
= 1.79 mole
Moles of 
= 1.20 mole
Moles of = 3.71 mole
Now we have to calculate the mole fraction of and gases.
and,
and,
Thus, the mole fraction of and gases are, 0.267, 0.179 and 0.554 respectively.
Now we have to calculate the partial pressure of and gases.
According to the Raoult's law,
where,
= partial pressure of gas
= total pressure of gas = 5.78 atm
= mole fraction of gas
and,
and,
Thus, the partial pressure of and gases are, 1.54, 1.03 and 3.20 atm respectively.
A chemical or physical change
