Answer : The value of ΔH° for this reaction is, 112.4 kJ/mol
Explanation :
According to the Van't Hoff equation,
[tex]\ln K=-(\frac{\Delta H^o}{R})(\frac{1}{T})+\frac{\Delta S^o}{R}[/tex]
The linear equation expression is:
y = mx + c
Plot of ln (K) versus (1/T) gives a straight line with a slope is equal to, [tex]\frac{-\Delta H^o}{R}[/tex]
As we are given that:
Slope = [tex]1.352\times 10^4K[/tex]
So,
[tex]Slope=\frac{-\Delta H^o}{R}=1.352\times 10^4K[/tex]
where,
R = 8.314 J/K.mol
[tex]\frac{-\Delta H^o}{R}=1.352\times 10^4K[/tex]
[tex]\Delta H^o=-(1.352\times 10^4K)\times R[/tex]
[tex]\Delta H^o=-(1.352\times 10^4K)\times (8.314J/K.mol)[/tex]
[tex]\Delta H^o=-112405.28J/mol=112.4kJ/mol[/tex]
Thus, the value of ΔH° for this reaction is, 112.4 kJ/mol
