Question

1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change that occurs during the process for your proof).
2. Work out the entropy change for the decomposition of mercuric oxide using mathematical and graphical arguments.

Answer 1

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

1) To show that heat flows spontaneously from high temperature to low temperature

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) Entropy change for Decomposition of mercuric oxide

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k