Answer
Verified
408.9k+ views
Hint: The measure of randomness or disordered distribution is known as entropy. The randomness is always higher in a gaseous state. More the number of gaseous molecules higher is the entropy. To solve this we must know the expression that gives the relation between free energy, entropy and enthalpy.
Formula Used:$\Delta {\text{S}}_{{\text{reaction}}}^ \circ = \Delta {{\text{S}}_{{\text{products}}}} - \Delta {{\text{S}}_{{\text{reactants}}}}$
$\Delta {{\text{G}}^ \circ } = \Delta {{\text{H}}^ \circ } - {\text{T}}\Delta {{\text{S}}^ \circ }$
Complete step-by-step answer:VWe are given the reaction as follows:
${\text{CO}}\left( {\text{g}} \right) + \dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
We know that the measure of randomness or disordered distribution is known as entropy.
We will first calculate the change in entropy of the reaction using the equation as follows:
$\Delta {\text{S}}_{{\text{reaction}}}^ \circ = \Delta {{\text{S}}_{{\text{products}}}} - \Delta {{\text{S}}_{{\text{reactants}}}}$
Where, $\Delta {\text{S}}_{{\text{reaction}}}^ \circ $ is the standard change in entropy of the reaction.
We are given the values of standard entropies as ${\text{S}}_{{\text{C}}{{\text{O}}_2}}^ \circ = 213.8{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$, ${\text{S}}_{{\text{CO}}}^ \circ = 197.9{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$, ${\text{S}}_{{{\text{O}}_2}}^ \circ = 205.8{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$. Thus,
$\Delta {\text{S}}_{{\text{reaction}}}^ \circ = {\text{213}}{\text{.8}} - \left( {197.9 + \dfrac{1}{2} \times 205.8} \right)$
$\Delta {\text{S}}_{{\text{reaction}}}^ \circ = - 87{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$
Thus, the standard change in entropy of the reaction is $ - 87{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} = - 87 \times {10^{ - 3}}{\text{ kJ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$.
We know the expression that gives the relation between free energy, entropy and enthalpy is as follows:
$\Delta {{\text{G}}^ \circ } = \Delta {{\text{H}}^ \circ } - {\text{T}}\Delta {{\text{S}}^ \circ }$
Where, $\Delta {{\text{G}}^ \circ }$ is the standard change in Gibb’s free energy,
$\Delta {{\text{H}}^ \circ }$ is the standard change in enthalpy,
${\text{T}}$ is the temperature,
$\Delta {{\text{S}}^ \circ }$ is the standard change in entropy.
We are given the standard values. At standard condition, ${\text{T}} = {25^ \circ }{\text{C}} = 298{\text{ K}}$. We are given that $\Delta {{\text{H}}^ \circ } = - 282.84{\text{ kJ}}$. Thus,
$\Delta {{\text{G}}^ \circ } = - 282.84{\text{ kJ}} - \left[ {{\text{298 K}}\left( { - 87 \times {{10}^{ - 3}}{\text{ kJ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}} \right)} \right]$
$\Delta {{\text{G}}^ \circ } = - 256.914{\text{ kJ}}$
Thus, $\Delta {{\text{G}}^ \circ }$ for the given reaction is $ - 256.91{\text{ kJ}}$.
Thus, the correct option is (3) $ - 256.91{\text{ kJ}}$.
Note:: We know that if the free energy change is negative then the reaction is spontaneous. If the free energy change is positive then the reaction is non-spontaneous. Here, $\Delta {{\text{G}}^ \circ } = - 256.914{\text{ kJ}}$ which is a negative value. Thus, the given reaction is a spontaneous reaction.
Formula Used:$\Delta {\text{S}}_{{\text{reaction}}}^ \circ = \Delta {{\text{S}}_{{\text{products}}}} - \Delta {{\text{S}}_{{\text{reactants}}}}$
$\Delta {{\text{G}}^ \circ } = \Delta {{\text{H}}^ \circ } - {\text{T}}\Delta {{\text{S}}^ \circ }$
Complete step-by-step answer:VWe are given the reaction as follows:
${\text{CO}}\left( {\text{g}} \right) + \dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
We know that the measure of randomness or disordered distribution is known as entropy.
We will first calculate the change in entropy of the reaction using the equation as follows:
$\Delta {\text{S}}_{{\text{reaction}}}^ \circ = \Delta {{\text{S}}_{{\text{products}}}} - \Delta {{\text{S}}_{{\text{reactants}}}}$
Where, $\Delta {\text{S}}_{{\text{reaction}}}^ \circ $ is the standard change in entropy of the reaction.
We are given the values of standard entropies as ${\text{S}}_{{\text{C}}{{\text{O}}_2}}^ \circ = 213.8{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$, ${\text{S}}_{{\text{CO}}}^ \circ = 197.9{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$, ${\text{S}}_{{{\text{O}}_2}}^ \circ = 205.8{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$. Thus,
$\Delta {\text{S}}_{{\text{reaction}}}^ \circ = {\text{213}}{\text{.8}} - \left( {197.9 + \dfrac{1}{2} \times 205.8} \right)$
$\Delta {\text{S}}_{{\text{reaction}}}^ \circ = - 87{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$
Thus, the standard change in entropy of the reaction is $ - 87{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} = - 87 \times {10^{ - 3}}{\text{ kJ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$.
We know the expression that gives the relation between free energy, entropy and enthalpy is as follows:
$\Delta {{\text{G}}^ \circ } = \Delta {{\text{H}}^ \circ } - {\text{T}}\Delta {{\text{S}}^ \circ }$
Where, $\Delta {{\text{G}}^ \circ }$ is the standard change in Gibb’s free energy,
$\Delta {{\text{H}}^ \circ }$ is the standard change in enthalpy,
${\text{T}}$ is the temperature,
$\Delta {{\text{S}}^ \circ }$ is the standard change in entropy.
We are given the standard values. At standard condition, ${\text{T}} = {25^ \circ }{\text{C}} = 298{\text{ K}}$. We are given that $\Delta {{\text{H}}^ \circ } = - 282.84{\text{ kJ}}$. Thus,
$\Delta {{\text{G}}^ \circ } = - 282.84{\text{ kJ}} - \left[ {{\text{298 K}}\left( { - 87 \times {{10}^{ - 3}}{\text{ kJ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}} \right)} \right]$
$\Delta {{\text{G}}^ \circ } = - 256.914{\text{ kJ}}$
Thus, $\Delta {{\text{G}}^ \circ }$ for the given reaction is $ - 256.91{\text{ kJ}}$.
Thus, the correct option is (3) $ - 256.91{\text{ kJ}}$.
Note:: We know that if the free energy change is negative then the reaction is spontaneous. If the free energy change is positive then the reaction is non-spontaneous. Here, $\Delta {{\text{G}}^ \circ } = - 256.914{\text{ kJ}}$ which is a negative value. Thus, the given reaction is a spontaneous reaction.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE