For the question
$A{g_2}O \rightleftharpoons 2A{g_{(s)}} + \dfrac{1}{2}{O_{2(g)}}$
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 $\left( {\Delta H = + 30.56kJmo{l^{ - 1}};\Delta S = + 0.066kJ{K^{ - 1}}mo{l^{ - 1}}} \right)$
Answer
585.3k+ views
Hint: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 $\Delta G = \Delta H - T\Delta S$, 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
$\Delta G = \Delta H - T\Delta S$
To get the temperature, modify the equation to
\[
T\Delta S = \Delta H - \Delta G \\
{\text{ }}T = \dfrac{{\Delta H - \Delta G}}{{\Delta S}} \\
\]
On substituting the given values, we get
\[
T = \dfrac{{30.56kJmo{l^{ - 1}} - 0kJmo{l^{ - 1}}}}{{0.066kJ{K^{ - 1}}mo{l^{ - 1}}}} \\
T = 463.03K \\
\]
2) 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.
3) 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.
Complete step by step answer:
1) To solve the first part, substitute the given values in this equation
$\Delta G = \Delta H - T\Delta S$
To get the temperature, modify the equation to
\[
T\Delta S = \Delta H - \Delta G \\
{\text{ }}T = \dfrac{{\Delta H - \Delta G}}{{\Delta S}} \\
\]
On substituting the given values, we get
\[
T = \dfrac{{30.56kJmo{l^{ - 1}} - 0kJmo{l^{ - 1}}}}{{0.066kJ{K^{ - 1}}mo{l^{ - 1}}}} \\
T = 463.03K \\
\]
2) 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.
3) 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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain the structure of megasporangium class 12 biology CBSE

What are the major means of transport Explain each class 12 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

