Question
Chloroform has $\Delta H_{\text { vaporization }}=29.2 \mathrm{kJ} / \mathrm{mol}$ and boils at$61.2^{\circ} \mathrm{C} .$ What is the value of $\Delta S_{\text { vaporization }}$ for chloroform?
Step 1
We can do this by adding 273.15 to the Celsius temperature. This is because the Kelvin scale starts at absolute zero, which is -273.15 degrees Celsius. So, we have: \[ T = 61.2^{\circ}C + 273.15 = 334.4 K \] Show more…
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Chloroform has $\Delta H_{\text {vaporization }}=29.2 \mathrm{~kJ} / \mathrm{mol}$ and boils at $61.2^{\circ} \mathrm{C} .$ What is the value of $\Delta S_{\text {vaporization }}$ for chloroform? (a) $87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ (b) $477.1 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ (c) $-87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ (d) $-477.1 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$
Chloroform $\left(\mathrm{CHCl}_{3}\right)$ has a normal boiling point of $61^{\circ} \mathrm{C}$ and an enthalpy of vaporization of 29.24 $\mathrm{kJ} / \mathrm{mol} .$ What are its values of $\Delta G_{\mathrm{vap}}$ and $\Delta S_{\mathrm{vap}}$ at $61^{\circ} \mathrm{C} ?$
Chloroform $\left(\mathrm{CHCl}_{3}\right)$ has $\Delta H_{\mathrm{vap}}=29.2 \mathrm{kJ} / \mathrm{mol}$ and $\Delta S_{\mathrm{vap}}=$ 87.5 $\mathrm{J} /(\mathrm{K} \cdot \mathrm{mol}) .$ What is the boiling point of chloroform in kelvin?
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