Respuesta :
Answer:
The reaction is not spontaneous in the forward direction, but in the reverse direction.
Explanation:
Step 1: Data given
H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol
Temperature = 25°C = 25+273 = 298 Kelvin
The initial pressures are:
pH2 = 3.10 atm
pI2 = 1.5 atm
pHI 1.75 atm
Step 2: Calculate ΔG
ΔG = ΔG° + RTln Q
with ΔG° = 2.60 kJ/mol
with R = 8.3145 J/K*mol
with T = 298 Kelvin
Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]
with pH2 = 3.10 atm
pI2 = 1.5 atm
pHI 1.75 atm
Q = (3.10²)/(1.5*1.75)
Q = 3.661
ΔG = ΔG° + RTln Q
ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)
ΔG =5815.43 J/mol = 5.815 kJ/mol
To be spontaneous, ΔG should be <0.
ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.
The reaction is not spontaneous in the forward direction, but in the reverse direction.
Given information:
H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol
Temperature = 25°C = 25+273 = 298 Kelvin
Initial pressure are pH2 = 3.10 atm, pI2 = 1.5 atm, and pHI 1.75 atm
calculation of energy:
Calculate ΔG
ΔG = ΔG° + RTln Q
Here
Here ΔG° = 2.60 kJ/mol
Here R = 8.3145 J/K*mol
Here T = 298 Kelvin
Now
Q = the reaction quotient
In this case [tex]Kp = [p(HI)]^2\div [p(H_2)] [p(I_2)][/tex]
Here pH2 = 3.10 atm
pI2 = 1.5 atm
pHI 1.75 atm
Now
[tex]Q = (3.10^2)\div (1.5\times 1.75)[/tex]
Q = 3.661
Now
ΔG = ΔG° + RTln Q
[tex]\Delta G = 2600 J/mol + 8.3145 J/K\times mol \times 298 K \times ln(3.661)[/tex]
ΔG =5815.43 J/mol
= 5.815 kJ/mol
To be spontaneous, ΔG should be <0.
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