Question

Consider the following processes and determine if the entropy change of the system is positive, negative, zero, or you cannot tell. Explain. (a) Liquid water is frozen to form ice. (b) 1 mol of oxygen and 2 mol of hydrogen react isothermally and completely to form 1 mol of water vapor. (c) 1 mol of oxygen and 2 mol of hydrogen react adiabatically and completely to form 1 mol of water vapor.

Answer 1

a. Liquid water is frozen to ice it means it liberated heat and constant its temperature and reduce randomness make a crystal form and we know that entropy is nothing but degree of randomness so we can say that here entropy is reduces and entropy change will become negative and also we know that entropy is the function of temperature and heat if system takes heat and temperature increases then obviously entropy increases but here heat release then entropy reduced and entropy change will be negative.

b. We know that in isothermal entropy change is R. ln(V2/V1)

V2 volume at condition 2

V1 volume at condition 1

So here 1mol Oxygen and 2 mol Hydrogen get reacts and form 1 mol Water it means here volume

V2<V1 so entropy change will be negative.

BUT HERE REACTION IS NOT CARRIED OUT BEACAUSE SO WE CAN'T SAY ABOUT ENTROPY CHANGE BEACAUSE REACTION IS NOT POSSIBLE STOCHIOMETRICALLY NOT TRUE.

H2 +1/2 O2 -> H2O

1mol hydorgen

1/2 mol oxygen form 1 mol water.

c. In reversible reaction adiabatic is isentropic means entropy will be constant but here also given 1 mol O2 and 2 mol Hydorgen gives 1 mol water which is not possible stoichemetrically as i say above.

SO ENTROPY CHANGE CAN NOT SAY IF THIS REACTION WILL BE OCCUR AS IT IS GIVEN.