![]() ![]() If ∆S is negative, then the negative signs (from the subtraction and the sign of ∆S) will cancel out, and so as T∆S gets bigger, ∆G will get more positive. Subtract the entropies of the reactants from the entropies of the products. T is always positive, so if ∆S is positive then a bigger T∆S will make ∆G more negative (since we subtract T∆S). The total entropy is two times 188.7 plus 213.6 which is 591 joules per Kelvin. As T increases, the T∆S component gets bigger. ∆H is still positive and ∆S is still whatever sign you figured out above. But the answer to that question says a lot. ![]() ![]() (1) in the introduction, now in terms of rates for a continuous process. For a heat transfer process between two temperature levels Ta T a and Tb T b the two parts (i) and (ii) are. Since ∆H and ∆S don't change significantly with temperature (given in the question), we can assume that they keep the same signs and values: i.e. Updated: What is Entropy Is it easier to make a mess or clean it up That might seem like an odd question, and a pretty easy one to answer. In a real (irreversible) process the change of entropy thus always is the sum of both, i.e. If ∆G is negative (from the question), is the reaction spontaneous or non-spontaneous?Ģ) Let's use ∆G = ∆H - T∆S again. From these values, we can know for certain whether ∆S is positive or negative (hint: remember that we are subtracting ∆G!).ġ) Knowing the sign of ∆G is enough to say whether the reaction is spontaneous or not under these conditions. Temperature is always positive (in Kelvin). We know (from the question) that ∆G is negative and that ∆H is positive. This looks like a homework question, so I'll give you some hints to get you on the riht path rather than answering directly.ģ) We know that ∆G = ∆H - T∆S. ![]()
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