Consider the following reaction for which the sign of enthalpy change is given.

Question

Consider the following reaction for which the sign of enthalpy change is given. Which of the reactors (I-V) are spontaneous at low temperatures only? Assume standard Cl2O (g) + 3/2 O2 (g) rightarrow 2ClO2 (g) delta Ho(+) 2SO2 (g) + O2 (g) rightarrow 2SO3 (g) delta Ho(-) 2NO2 (g) rightarrow N2O4 (g) delta Ho(-) 6CO2 (g) + 6H2O (g) rightarrow C6H12O6 (g) + 6O2 9g) delta Ho(+) HCL (g) + NH3 (g) rightarrow NH4Cl (s) delta Ho(-) Please enter your answers separated with commas,do not use the word and; use no spaces; use Roman numerals and enter your answer in numerical order. If the answer is none of the equations listed enter the word none. For example, if the answer is equations 11 and III enter 11,111

Explanation / Answer

delta G = delta H - T (delta S)

if delta G is -ve, then reaction is spontaneous, otherwise non-spontaneous

We know the values of Delta H of the reactions

so Delta S values will define whether reaction will be spontaneous or not

Delta S is governed by difference of Number of gases of products and reactants

If delta ng is positive i.e. no. of moles of products is greater than no. of moles of gases in reactant

implies delta S is positive;

for reaction to be spontaneous delta S should be +ve(mainly)

1. delta H is +ve;

delta ng = -1/2

Therefore delta G is + ve ; non-spontaneous

2.Delta H is -ve

delta ng = -ve ;

since T is low, so delta G is -ve because delta H > T * delta S

reaction is spontaneous

3.delta H is -ve;

delta ng = -1

since T is low, so delta G is -ve because delta H > T * delta S

reaction is spontaneous

4.delta H is +ve;

delta ng=-5;Therefore delta G is +ve;

reaction is non-spontaneous

5. delta H is -ve;

delta ng= -2;

delta S is also -ve

since T is low, so delta G is -ve because delta H > T * delta S

reaction is spontaneous

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