Solveeit Logo
Login

Question

Question: For the question \(A{g_2}O \rightleftharpoons 2A{g_{(s)}} + \dfrac{1}{2}{O_{2(g)}}\) Calculate ...

For the question
Ag2O⇌2Ag(s)+12O2(g)A{g_2}O \rightleftharpoons 2A{g_{(s)}} + \dfrac{1}{2}{O_{2(g)}}
Calculate the temperature at which free change is zero. At a temperature lower than this, predict whether the forward of the reverse reaction will be formed. Give reason (ΔH=+30.56kJmol−1;ΔS=+0.066kJK−1mol−1)\left( {\Delta H = + 30.56kJmo{l^{ - 1}};\Delta S = + 0.066kJ{K^{ - 1}}mo{l^{ - 1}}} \right)

Explanation

Solution

There are two parts in the question, one to calculate the temperature at which the free energy is zero, other to say whether the reaction goes forward or backward. In the question they have given values of change in enthalpy and entropy, if we remember there is formula which relates free energy change, temperature, enthalpy change and entropy change ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, using this we can calculate the temperature. To solve the other part, now see the enthalpy change value, its positive value which says that the reaction is endothermic using le chatelier’s principle we can figure out the answer.

Complete step by step answer:

  1. To solve the first part, substitute the given values in this equation
    ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
    To get the temperature, modify the equation to
TΔS=ΔH−ΔG  T=ΔH−ΔGΔS  T\Delta S = \Delta H - \Delta G \\\ {\text{ }}T = \dfrac{{\Delta H - \Delta G}}{{\Delta S}} \\\

On substituting the given values, we get

T=30.56kJmol−1−0kJmol−10.066kJK−1mol−1 T=463.03K  T = \dfrac{{30.56kJmo{l^{ - 1}} - 0kJmo{l^{ - 1}}}}{{0.066kJ{K^{ - 1}}mo{l^{ - 1}}}} \\\ T = 463.03K \\\
  1. According to Le chatelier’s principle, the temperature dependence of the equilibrium constant is in turn dependent on the sign of the enthalpy change.
    -The equilibrium constant for the exothermic reaction reduces as the temperature is put up.
    -The equilibrium constant for the endothermic reaction increases as the temperature is put up.
  2. Here the given reaction is endothermic, and the question says to decrease the temperature of the system which impairs the equilibrium constant of the system. After we decrease the temperature, the system tries to counteract the change by carrying out the backward reaction, so that the equilibrium constant is maintained.

Note: Le chatelier’s principle states that any change in the factor that determines the equilibrium conditions of a system will lead the system to change such as to counteract the effect of the change. We should keep all the points of this law while dealing with a reaction at equilibrium conditions.