
Find ${{\left( \text{ }\!\!\Delta\!\!\text{ S} \right)}_{\text{universe}}}$ (in Joule/mole/K) for a chemical reaction at 300K if $\text{ }\!\!\Delta\!\!\text{ H}_{\text{300K}}^{\text{0}}\text{=75kJ/mol}$.$\text{ }\!\!\Delta\!\!\text{ S}_{\text{300K}}^{\text{0}}\text{=300J/K}$
Answer
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Hint Entropy of the universe can be calculated using the formulae,
${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{universe}}}\text{=}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{system}}}\text{+}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{surrounding}}}$
Entropy of the system can be calculated using the formulae,
$\text{ }\!\!\Delta\!\!\text{ S=}\dfrac{\text{q}}{\text{T}}$
$\text{ }\!\!\Delta\!\!\text{ S=}\dfrac{\text{q}}{\text{T}}\text{=}\dfrac{\text{ }\!\!\Delta\!\!\text{ H}}{\text{T}}$, here q is the amount of heat for the temperature change and T is the temperature.
Complete step by step solution:
Entropy is the measure of randomness of the system. According to the second law of thermodynamics entropy of the Universe is constantly increasing and it is the summation of entropy of a system or say entropy of reaction and entropy of surrounding.
And for spontaneous reaction, entropy will be maximum and at equilibrium the change in entropy will be zero i.e. $\text{ }\!\!\Delta\!\!\text{ S=0}$
And we can write entropy as the amount of heat for the temperature change which will be equal to the change in enthalpy with temperature.
And hence we can write the equation as,
$\text{ }\!\!\Delta\!\!\text{ S=}\dfrac{\text{q}}{\text{T}}\text{=}\dfrac{\text{ }\!\!\Delta\!\!\text{ H}}{\text{T}}$
Now let’s check the data provided in the question,
It is given that,
$\text{ }\!\!\Delta\!\!\text{ H}_{\text{300K}}^{\text{0}}\text{=75kJ/mol}$, which is change in enthalpy of the system or reaction.
Temperature (T) =300 K
$\text{ }\!\!\Delta\!\!\text{ S}_{\text{300K}}^{\text{0}}\text{=300J/K}$
Enthalpy is given in kilojoules, so it should be converted into Joules.
And the equation for entropy of system is,
$\text{ }\!\!\Delta\!\!\text{ }{{\text{S}}_{\text{reaction}}}\text{=}\dfrac{\text{ }\!\!\Delta\!\!\text{ H}}{\text{T}}\text{=}\dfrac{\text{75 }\!\!\times\!\!\text{ 1000}}{\text{300}}$
$\text{ }\!\!\Delta\!\!\text{ }{{\text{S}}_{\text{reaction}}}\text{=250J/K}$
${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{universe}}}\text{=}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{system}}}\text{+}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{surrounding}}}$
${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{universe}}}\text{=250+300=550J/K}$
Here the entropy is positive, so it is a spontaneous reaction.
Note: While doing such problems the units should be taken in consideration. As in this case enthalpy was given in Kilojoules and it has to be converted into joules.
And attention should be given for signs as for non-spontaneous reactions, the entropy will be negative.
And enthalpy also we have both negative and positive values, so it should be taken care off.
${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{universe}}}\text{=}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{system}}}\text{+}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{surrounding}}}$
Entropy of the system can be calculated using the formulae,
$\text{ }\!\!\Delta\!\!\text{ S=}\dfrac{\text{q}}{\text{T}}$
$\text{ }\!\!\Delta\!\!\text{ S=}\dfrac{\text{q}}{\text{T}}\text{=}\dfrac{\text{ }\!\!\Delta\!\!\text{ H}}{\text{T}}$, here q is the amount of heat for the temperature change and T is the temperature.
Complete step by step solution:
Entropy is the measure of randomness of the system. According to the second law of thermodynamics entropy of the Universe is constantly increasing and it is the summation of entropy of a system or say entropy of reaction and entropy of surrounding.
And for spontaneous reaction, entropy will be maximum and at equilibrium the change in entropy will be zero i.e. $\text{ }\!\!\Delta\!\!\text{ S=0}$
And we can write entropy as the amount of heat for the temperature change which will be equal to the change in enthalpy with temperature.
And hence we can write the equation as,
$\text{ }\!\!\Delta\!\!\text{ S=}\dfrac{\text{q}}{\text{T}}\text{=}\dfrac{\text{ }\!\!\Delta\!\!\text{ H}}{\text{T}}$
Now let’s check the data provided in the question,
It is given that,
$\text{ }\!\!\Delta\!\!\text{ H}_{\text{300K}}^{\text{0}}\text{=75kJ/mol}$, which is change in enthalpy of the system or reaction.
Temperature (T) =300 K
$\text{ }\!\!\Delta\!\!\text{ S}_{\text{300K}}^{\text{0}}\text{=300J/K}$
Enthalpy is given in kilojoules, so it should be converted into Joules.
And the equation for entropy of system is,
$\text{ }\!\!\Delta\!\!\text{ }{{\text{S}}_{\text{reaction}}}\text{=}\dfrac{\text{ }\!\!\Delta\!\!\text{ H}}{\text{T}}\text{=}\dfrac{\text{75 }\!\!\times\!\!\text{ 1000}}{\text{300}}$
$\text{ }\!\!\Delta\!\!\text{ }{{\text{S}}_{\text{reaction}}}\text{=250J/K}$
${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{universe}}}\text{=}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{system}}}\text{+}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{surrounding}}}$
${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{universe}}}\text{=250+300=550J/K}$
Here the entropy is positive, so it is a spontaneous reaction.
Note: While doing such problems the units should be taken in consideration. As in this case enthalpy was given in Kilojoules and it has to be converted into joules.
And attention should be given for signs as for non-spontaneous reactions, the entropy will be negative.
And enthalpy also we have both negative and positive values, so it should be taken care off.
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