Use your equations here. Delta g = delta h - t*Delta S
If t is zero, delta g is delta h. Since it's nonspon, delta g is greater than zero, so delta h is greater than zero. Check with the converse. If t is very high, delta g is going to be very negative, regardless of what sign h has. Endothermic makes sense here. Hope that helps!
Endothermic or exothermic depends on enthalpy, spontanaeity depends on delta g. If t is high, delta g = delta h - very large number
Delta h = delta g + very large number
Delta h > 0 or endothermic
If t =0, delta g = delta h. If the reaction is spontaneous then delta g < 0, if it's not, delta g > 0. Since we're given it's not spontaneous at low temperatures delta g > 0, and delta h is also greater than zero. Which is endothermic.
Thanks, I can see it now , so we can we have a spont and endothermic at the same time, if delta H is negative then it is automatically spont and exothermic
If Delta H is positive then it can be spon or Non spon depending on how large T is.