Question

Predict whether S for each reaction would be greater than zero, less than zero, or too close to zero to decide. Clear All H2(g) + F2(g)2HF(g) 2SO3(g)2SO2(g) + O2(g) CO2(g) + H2(g)CO(g) + H2O(g) 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) 2H2O2(l)2H2O(l) + O2(g) S > 0 S < 0 too close to decide

Answer 1

Entropy of gas > Entropy of liquid > Entropy of solid

1) H2(g) + F2(g) --> 2HF(g)

Product side with 2 gaseous moles

Reactant side with (1+1) 2 gaseous moles. Thus, Delta S will be close to zero. But in actual, Delta S is 14 J/K.

2) 2SO3(g) --> 2SO2(g) + O2(g)

Product side has 3 gaseous moles and reactant side 2 gaseous moles, thus Delta S is positive i.e. greater than 0.

3) CO2(g) + H2(g) --> CO(g) + H2O(g)

Both side have equal moles of gases, So Delta S should be close to 0 but in actual Delta S is 42 which is greater than zero.

4) 4HCl(g) + O2(g) --> 2H2O(g) + 2Cl2(g)

Product side has lesser gaseous moles than reactant side, thus Delta S is negative i.e. less than 0.

5) 2H2O2(l) --> 2H2O(l) + O2 (g)

Delta S is positive i.e. greater than zero as product side has 1 gaseous and 2 liquid moles while product side has only 2 liquid moles

Let me know if any doubt.