Which reaction below should have deltaSdegree > 0? 2H2(g) + 02(g) rightarrow 2H2

Question

Which reaction below should have deltaSdegree > 0? 2H2(g) + 02(g) rightarrow 2H2O(g) 2N02(g) rightarrow N2O4(g) H+ (aq) + F(aq) rightarrow HF(aq) Baf_2 (s) rightarrow Ba2 + (aq) + 2f (aq) 2Hg(l) + 02(g) rightarrow 2HgO(s) Which equation represents a reaction for which deltasdegree is negative at 25degreeC? Some standard entropies (it 25degreeC In J/mol K) are given: diamond 2 43. 02(g) 205 0 CO(g) 197 9. The value of 5+ for the reaction. Mcrcury freeies at -38 9degreeC. and its enthalpy of fusion Is 2.33 kJ/mol. Calculate the value of deltaS for the freezing of 1.00 mol of Hg(l) at = 38.9degreeC Assuming A Hand AS do not vary with temperatur at what temperature will the reaction shown below become spontaneous?

Explanation / Answer

Solution :-

Q1) The reaction shown in the option d is the reaction which will have the delta S > 0 because the solid is changed to aqueous ions which increases the entropy

So the answer is option d

Q2) In the reaction b the 3 gas moles on the reactant sides are changed to liquid water so the change from the gas to liquid is the decrease in entropy so the delta S value of the reaction will be negative so the answer is option b.

Q3)

2C(diamond) + O2(g) ---- > 2CO(g)

Delta S rxn = sum of delta S product - sum of delta S reactant

                  =[2*CO] –[(2* C)+(1*O2)]

                  = [197.9*2] – [(2.43*2)+(205.0*1)]

               = 188.37 J/K

Q4) Delta H fusion = 2.33 kJ /mol * 1000 J / 1 kJ= 23300 J/mol

T= -38.9 C +273 = 234.1 K

Delta S= Delta H / T

          = 23300 J per mol / 234.1K

           = 9.95 J/mol

Q5) Delta H= 131.3 kJ

Delta S = 133.6 J/K * 1 kJ/1000 J = 0.1336 kJ/K

To become the reaction spontaneous the delta G needs to be zero or less than zero

So

T= Delta H / delta S

= 131.3 kJ / 0.1336 kJ/mol

= 983 K

983 K – 273 = 710 C

So the reaction will become spontaneous at above 710 degree C

Q6) The delta H value is positive and delta S value is also positive

So the reaction is spontaneous at high temperature because the positive delta H means the reaction is endothermic.

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