Without doing a calculation, predict whether the entropy change will be positive
ID: 1055949 • Letter: W
Question
Without doing a calculation, predict whether the entropy change will be positive or negative when each reaction occurs in the direction it is written, For the reaction PbO(s) + C(s) rightarrow Pb(s) + CO(g) Delta H degree= +106.8 kJ and Delta S degree = + 188.0 J/K. What is the value of G degree at 298 K? Would this reaction occur spontaneously? From the values given for Delta H degree and Delta S degree, predict if each reaction would be spontaneous at 298 K and if not, at what temperature the reaction would become spontaneous.Explanation / Answer
Answer 1(a) Entropy decreases since there 3 moles gaseous reactants while only 1 mole of gaseous product.
1 (b) Entropy decreses since 3 moles of gaseous reactants while onle 1 mole of gaseous product.
1(c) Entropy increases as there are no gaseous reactant while there is one gaseous product.
Answer 2. dG0 = dH0 - TdS0
dG0 = 106.8 kJ (1000 J/ 1kJ) - 298 K x (+188 J/K)
dG0 = 106800 J - 56024 J = +50776 J
Since free energy change dG0 value is positive this reaction will be non-spontaneous.
Answer 3 (a)
dG0 = dH0 - TdS0
dG0 = - 844 kJ (1000 J/ 1kJ) - 298 K x (-165 J/K)
dG0 = - 844000 J + 49170 J = -794830 J = -794.830 kJ
Since free energy change dG0 is negative for this process this reaction will be spontaneous.
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