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.
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
