Which aldehyde is an intermediate in the reduction of ethyl benzoate with lithium aluminum hydride?
1 answer:
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Use the Clausius-Clapeyron equation...
<span>Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.
</span>![ln(P_1/P_2) = \frac{-\delta H_{vap}}{R} \times [\frac{1}{T_1} - \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28P_1%2FP_2%29%20%3D%20%5Cfrac%7B-%5Cdelta%20H_%7Bvap%7D%7D%7BR%7D%20%20%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
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</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K)
</span>
------ some algebra goes here -----
<span>T = 349.99K ...... or ...... 76.8C </span>
<span>Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.
</span>
</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K)
</span>
------ some algebra goes here -----
<span>T = 349.99K ...... or ...... 76.8C </span>