How many milliliters of 2.5M HCI are required to exactly neutralize 15 milliliters of 5.0 M NaOH
1 answer:
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Answer:
The boiling point is 308.27 K (35.27°C)
Explanation:
The chemical reaction for the boiling of titanium tetrachloride is shown below:
Ti ⇒ Ti
ΔH° (Ti
) = -804.2 kJ/mol
ΔH° (Ti
) = -763.2 kJ/mol
Therefore,
ΔH° = ΔH°
(Ti
) - ΔH°
(Ti
) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol
Similarly,
s°(Ti) = 221.9 J/(mol*K)
s°(Ti) = 354.9 J/(mol*K)
Therefore,
s° = s° (Ti) - s°(Ti
) = 354.9 - 221.9 = 133 J/(mol*K)
Thus, T = ΔH° /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C
Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.