Respuesta :
Answer: The value of [tex]\Delta S^o[/tex] for the surrounding when given amount of CO gas is reacted is 197.77 J/K
Explanation:
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]2CO(g)+2NO(g)\rightarrow 2CO_2(g)+N_2(g)[/tex]
The equation for the entropy change of the above reaction is:
[tex]\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})+(1\times \Delta S^o_{(N_2(g))})]-[(2\times \Delta S^o_{(CO(g))})+(2\times \Delta S^o_{(NO(g))})][/tex]
We are given:
[tex]\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(N_2(g))}=191.61J/K.mol\\\Delta S^o_{(CO(g))}=197.67J/K.mol\\\Delta S^o_{(NO(g))}=210.76J/K.mol[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(2\times (213.74))+(1\times (191.61))]-[(2\times (197.67))+(2\times (210.76))]\\\\\Delta S^o_{rxn}=-197.77/K[/tex]
Entropy change of the surrounding = - (Entropy change of the system) = -(-197.77) J/K = 197.77 J/K
We are given:
Moles of CO gas reacted = 2.00 moles
By Stoichiometry of the reaction:
When 2 mole of CO gas is reacted, the entropy change of the surrounding will be 197.77 J/K
So, when 2.00 moles of CO gas is reacted, the entropy change of the surrounding will be = [tex]\frac{197.77}{2}\times 2.00=197.77J/K[/tex]
Hence, the value of [tex]\Delta S^o[/tex] for the surrounding when given amount of CO gas is reacted is 197.77 J/K
