Answer:
Part 1: - 1.091 x 10⁴ J/mol.
Part 2: - 1.137 x 10⁴ J/mol.
Explanation:
Part 1: At standard conditions:
At standard conditions Kp= 81.9.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.
Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.
For the reaction:
I₂(g) + Cl₂(g) ⇌ 2ICl(g).
Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.
<u>Given information:</u>
Concentration of HCl = 0.035 M
<u>To determine:</u>
pH of the solution
<u>Explanation:</u>
Hydrochloric acid, HCl is a strong acid. It will completely dissociate to give H+ and Cl- ions
HCl → H+ + Cl-
Hence the concentration of H+ = Cl- = 0.035M
Now, pH measures the strength of H+ ions in a given solution. It is expressed as:
pH = -log[H+]
pH (HCl) = -log(0.035) = 1.46
Ans: pH of 0.035M HCl is 1.46
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