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iogann1982 [59]
3 years ago
9

Find the boiling point temperature at 760 torr of an isomer of octane, c8h18, if its enthalpy of vaporization is 38,210 j mol-1

and its vapor pressure at 110.0°c is 638.43 torr.
Chemistry
1 answer:
mixas84 [53]3 years ago
7 0

According to Clausius-Clayperon equation,

ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} }   )

P_{1}is the vapor pressure at boiling point = 760 torr

P_{2} is the vapor pressure at T_{2} =638.43 torr

TemperatureT_{2} = 110.0^{0}C + 273 = 383 K

ΔH_{vap} = 38210 J/mol

Plugging in the values, we get

ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} }   )

ln\frac{760 torr}{638.43 torr} =\frac{38210 J/mol}{8.314 J/(mol.K)} (\frac{1}{383 K}-\frac{1}{T_{1} }

T_{1} = 389 K

Therefore, the boiling point of octane = 389 K - 273 = 116^{0}C

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<h3>Balanced equation </h3>

P₄ + 6Cl₂ → 4PCl₃

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<h3>SUMMARY</h3>

From the balanced equation above,

124 g of P₄ reacted to produce 550 g of PCl₃

<h3>How to determine the theoretical yield </h3>

From the balanced equation above,

124 g of P₄ reacted to produce 550 g of PCl₃

Therefore,

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<h3>How to determine the percentage yield </h3>
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  • Theoretical yield of PCl₃ = 350.9 g
  • Percentage yield =?

Percentage yield = (Actual /Theoretical) × 100

Percentage yield = (262.6 / 350.9) × 100

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