Question

The combustion of acetylene C2H2 is a spontaneous reaction given by the equation: 2C2H2 (g) + SO2(g) --> 4CO2 (g) + 2 H2O(l)

As expected for a combustion the a combustion the reaction is exothermic.

a.) What is the sign of Delta H?

b.) What do you expect for the sign of Delta S?

c.) Explain the spontaneity of the reaction in terms of the enthalpy and entropy changes.

Please show work and explain

Answer 1

Answer – We are given, combustion of acetylene C₂H₂ is a spontaneous reaction-

2C₂H₂ (g) + SO₂(g) ----> 4CO₂ (g) + 2 H₂O(l)

a)We know the combustion reaction are exothermic and when reaction is exothermic means there is released heat to surrounding and change in enthalpy gets negative sign.

So sign for the   Delta H is negative.

b) This is the spontaneous reaction at room temp and we know deltaH is negative then we know form the spontaneous reaction, we must need deltaH is negative and delta S is also negative, since there is converting of gases to liquid and degree of randomness decrease. So at the low temp this reaction gets favored.

So sign of Delta S is negative.

c) So for the spontaneous reaction we must deltaH in negative and deltaS also negative and it is favored at only low temp. At high temp this reaction become non- spontaneous reaction and gets reversed. So as the temperature low there is more.