Calculate the entropy change in surroundings when $1{{mol}}$ of ${{{H}}_2}{{{O}}_{\left( {{l}} \right)}}$ is formed under the standard conditions at $298{{K}}$. Given ${\Delta _{{f}}}{{{H}}^ \circ } = - 286{{kJ}}.{{mo}}{{{l}}^{ - 1}}$
Answer
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Hint: Enthalpy is the total kinetic energy and potential energy of a system at constant pressure. Enthalpy change is the heat change of a system. While entropy is the measure of disorder or randomness. Entropy of the universe is constantly increasing.
Complete step by step answer:
Standard enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its elements under standard condition. It is the difference of enthalpy change of products and the enthalpy change of reactants.
Entropy increases if the system absorbs heat, otherwise entropy decreases. Entropy is always non-conservative, unlike energy.
The reaction is given below:
${{{H}}_{2\left( {{g}} \right)}} + \dfrac{1}{2}{{{O}}_{2\left( {{g}} \right)}} \to {{{H}}_2}{{{O}}_{\left( {{l}} \right)}}$
When one mole of water is formed, $286{{kJ}}$ is released at $298{{K}}$. The same heat is absorbed by the surroundings.
i.e. The amount of heat, ${{{q}}_{{s}}} = + 286{{kJ}}{{.mo}}{{{l}}^{ - 1}}$, Temperature, ${{T = 298K}}$
Entropy can be calculated by dividing the amount of heat absorbed by surroundings by the temperature.
i.e. $\Delta {{{S}}_{{s}}} = \dfrac{{{{{q}}_{{s}}}}}{{{T}}}$
Substituting the values, we get
\[\Delta {{{S}}_{{s}}} = \dfrac{{286{{kJ}}.{{mo}}{{{l}}^{ - 1}}}}{{298{{K}}}} = 0.96{{kJ}}.{{{K}}^{ - 1}}.{{mo}}{{{l}}^{ - 1}}\]
So the entropy change in the surroundings is \[0.96{{kJ}}.{{{K}}^{ - 1}}.{{mo}}{{{l}}^{ - 1}}\].
Additional information:
The entropy of the universe tends to be maximum. It has fixed values at fixed thermodynamic states. Hence the entropy change is determined by the initial and final state. Entropy change can be calculated by dividing heat amount by temperature. This is applicable only for a reversible process.
Note: Entropy shows the possibility of conversion of that into work. More disordered the system is then the entropy is also more. Entropy increases in solid-liquid conversion. While entropy decreases in liquid-solid and gas-liquid conversion.
Complete step by step answer:
Standard enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its elements under standard condition. It is the difference of enthalpy change of products and the enthalpy change of reactants.
Entropy increases if the system absorbs heat, otherwise entropy decreases. Entropy is always non-conservative, unlike energy.
The reaction is given below:
${{{H}}_{2\left( {{g}} \right)}} + \dfrac{1}{2}{{{O}}_{2\left( {{g}} \right)}} \to {{{H}}_2}{{{O}}_{\left( {{l}} \right)}}$
When one mole of water is formed, $286{{kJ}}$ is released at $298{{K}}$. The same heat is absorbed by the surroundings.
i.e. The amount of heat, ${{{q}}_{{s}}} = + 286{{kJ}}{{.mo}}{{{l}}^{ - 1}}$, Temperature, ${{T = 298K}}$
Entropy can be calculated by dividing the amount of heat absorbed by surroundings by the temperature.
i.e. $\Delta {{{S}}_{{s}}} = \dfrac{{{{{q}}_{{s}}}}}{{{T}}}$
Substituting the values, we get
\[\Delta {{{S}}_{{s}}} = \dfrac{{286{{kJ}}.{{mo}}{{{l}}^{ - 1}}}}{{298{{K}}}} = 0.96{{kJ}}.{{{K}}^{ - 1}}.{{mo}}{{{l}}^{ - 1}}\]
So the entropy change in the surroundings is \[0.96{{kJ}}.{{{K}}^{ - 1}}.{{mo}}{{{l}}^{ - 1}}\].
Additional information:
The entropy of the universe tends to be maximum. It has fixed values at fixed thermodynamic states. Hence the entropy change is determined by the initial and final state. Entropy change can be calculated by dividing heat amount by temperature. This is applicable only for a reversible process.
Note: Entropy shows the possibility of conversion of that into work. More disordered the system is then the entropy is also more. Entropy increases in solid-liquid conversion. While entropy decreases in liquid-solid and gas-liquid conversion.
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