Answer:
Explanation:
Answer:
The reaction is not spontaneous in the forward direction, but in the reverse direction.
Explanation:
<u>Step 1: </u>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
<u>Step 2</u>: 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 percentage of Chromium in Chromium Oxide is calculated as follow,
Step 1: Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
2(51.99) + 3(16) = 103.98 + 48 = 151.98 u
Step 2: Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
=

× 100
=
68.41 %
So, the %age composition of chromium in chromium oxide is
68.41 %.
Answer:
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