Without doing any calculations, predict the sign of
ΔSsys for each of the following processes:
1C(s) +
2H2(g) →
CH4(g)
2 4HF(g) +
O2(g) → 2H2O(l) +
2F2(g)
3 The freezing of
C3H6
4 The decomposition of
Zn3(PO4)2 into its constituent
elements
First to all, the entropy of elements or diatomic elements, is zero. So, with this and with the data of the entropy of each compound, you can predict the sign of the entropy.
In the first case, carbon would be zero, so entropy of system will be the entropy of the CH4 only. So it should be positive.
In the second case, it would be positive too, because you will not taking account the entropy of O2 which is zero.
In the third case, when you are freezing you are actually absorbing energy instead of releasing, so it should be negative.
In the fourth case, descomposition requires heat to be released, so it should be positive.
Hope this helps.
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