(9am to 6pm)

Ask Questions, Get Answers

Want help in doing your homework? We will solve it for you. Click to know more.

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} $

1 Answer

Need homework help? Click here.
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

Related questions