A) highly spontaneous at ordinary temperature
B) one with negligible enthalpy change
C) endothermic
D) exothermic
Correct Answer: D
Solution :
Variation of\[{{K}_{eq}}\]with temperature T is given by van't Hoff equation. \[\log {{K}_{eq}}=-\frac{\Delta {{H}^{o}}}{2.303RT}+\frac{\Delta {{S}^{o}}}{R}\] A B Slope of the given line is +ve indicating that term A is positive thus\[\Delta {{H}^{o}}\]is -ve. Thus reaction is exothermic.You need to login to perform this action.
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