A particular reaction at constant pressure is spontaneous at 390K. The enthalpy change for this reaction is +23.7kJ. What can you conclude about the sign and magnitude of ΔS for this reaction?a. smallb. largec. + smalld. + largee. 0.0

+ small

Explanation:

The entropy is obtained from;

∆S= ∆H/T

Where;

∆S= entropy of the system

∆H= enthalpy if the system = +23.7 KJ

T= absolute temperature of the system = 390 K

∆S= 23.7 ×10^3/390 = 60.8 JK^-

There is a small positive change in entropy.

b) nuclear ive learned that in a pedtrodic table

the answer your looking for is c).