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For the vaporization process of HI, the molar entropy and molar enthalpy are respectively, $89.0\;J \;K^{-1}\; mol^{-1}$ and $21.16\; kJ\; mol^{-1}$. The temperature at which the value of $\Delta G$ becomes zero for this process is:

$\begin{array}{1 1} 237800\;k \\ 273.8\;^{\circ} C \\ 35.2\;K \\ -35.2 ^{\circ} C \end{array} $

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Solution :
$\Delta _{vap} H= 21.16\;kJ\;mol^{-1}=21.16 \times 10^3 J/mol$
$\Delta _{vap}S =89.0\;J k^{-1} mol^{-1}$
$T = \Delta H/ \Delta S = 21.16x10^-3Jmol^-1/89Jmol^-1K^-1$
$\qquad= 237.8\;K$
$\qquad= (237.8-273)^{\circ}C = -35.2^{\circ}C$
Answer : $-35.2 ^{\circ} C$
answered Jan 18, 2016 by meena.p

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