Question

Calculate $\text{ }\Delta {{\text{S}}_{\text{rxn}}}\text{ }$ (in $\text{ J/K }$ ) for the chemical reaction: $\text{ C(graphite) + 2}{{\text{H}}_{\text{2}}}(g)\text{ }\to \text{ C}{{\text{H}}_{\text{4}}}\text{ }(g)\text{ }\Delta \text{H}_{\text{300}}^{\text{o}}\text{ = }-\text{75}\text{.0 kJ }$
The standard entropies of C (graphite), $\text{ }{{\text{H}}_{\text{2}}}(g)\text{ }$ and $\text{ C}{{\text{H}}_{\text{4}}}(g)\text{ }$ are $\text{ 6}\text{.0 }$ , $\text{ 130}\text{.6 }$ and $\text{ 186}\text{.2 J/K mol }$ respectively.
A) 169
B) 167
C) 170
D) 165

Answer 1

The change in entropy $\text{ }\Delta {{\text{S}}^{\text{o}}}\text{ }$for a chemical reaction from the values of the entropy of the reactant and the product at $\text{ 298 K }$ .for a general reaction given below,
$\text{ }a\text{A + }b\text{B + }............\to \text{ }l\text{L + }m\text{M +}............\text{ }$
The standard entropy change for the reaction is given by,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{S}}^{\text{o}}}\text{ = }\left[ l\text{S}_{\text{L}}^{\text{o}}\text{+}m\text{S}_{\text{M}}^{\text{o}}\text{+}.......... \right]-\left[ a\text{S}_{\text{A}}^{\text{o}}\text{+ }b\text{S}_{\text{B}}^{\text{o}}\text{+}.......... \right]\text{ }$
Or it can be also written as,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{S}}^{\text{o}}}\text{ = }\sum{\text{S}_{\text{products}}^{\text{o}}}-\text{ }\sum{\text{S}_{\text{reactant}}^{\text{o}}}\text{ }$
The entropy change of the surrounding is needed to be considered. It is equal to,
$\text{ }\Delta {{\text{S}}_{\text{surrounding}}}\text{ = }\dfrac{-\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{sys}}}{{{\text{T}}_{surrounding}}}\text{ }$

Complete step by step answer:
We are provided the following data: The graphite undergoes the hydrogenation reaction and forms the methane .reaction between the graphite and the hydrogen gas is as depicted below,
$\text{ C(graphite) + 2}{{\text{H}}_{\text{2}}}(g)\text{ }\to \text{ C}{{\text{H}}_{\text{4}}}\text{ }(g)\text{ }$
Enthalpy of reaction at $\text{ 300 K }$ is $\text{ }\Delta \text{H}_{\text{300}}^{\text{o}}\text{ = }-\text{75}\text{.0 kJ }$
Entropies for the:
C (graphite) is $\text{ S}_{\text{C(graphite)}}^{\text{o}}\text{ = 6}\text{.0 J}{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\text{ }$
Hydrogen gas is $\text{ S}_{{{\text{H}}_{\text{2}}}}^{\text{o}}\text{ = 130}\text{.6 J}{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\text{ }$
Methane gas is $\text{ S}_{\text{C}{{\text{H}}_{4}}}^{\text{o}}\text{ = 186}\text{.2 J}{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\text{ }$
We are interested in calculating the change in entropy of a reaction. We will solve this problem in two parts.

Part A) standard entropy of chemical reaction: The reaction is as shown below,
$\text{ C(graphite) + 2}{{\text{H}}_{\text{2}}}(g)\text{ }\to \text{ C}{{\text{H}}_{\text{4}}}\text{ }(g)\text{ }$
Let's first calculate the total standard entropy change for the reactant.
$\text{ }\sum{\text{S}_{\text{reactant}}^{\text{o}}}\text{ = S}_{\text{C(graphite)}}^{\text{o}}\text{ + 2S}_{{{\text{H}}_{\text{2}}}}^{\text{o}}\text{ }$
Substitute the values .we have,
$\text{ }\sum{\text{S}_{\text{reactant}}^{\text{o}}}\text{ = 6 + 2}\left( 130.6 \right)\text{ = 267}\text{.2 J}{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\text{ }$
The reaction involves one product thus total entropy of the reaction is given as,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{S}}^{\text{o}}}\text{ =}\left( 186.2 \right)-(267.2)\text{ = }-81\text{ J }{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\text{ }$
Part B) entropy change for surrounding: The heat of formation reaction or the enthalpy of formation of reaction is given as$\text{ }\Delta \text{H}_{\text{300}}^{\text{o}}\text{ = }-\text{75}\text{.0 kJ }$at the $\text{ 300 K }$ temperature. The entropy change of the surrounding from the enthalpy of reaction is calculated from the change in enthalpy at the absolute temperature. Thus entropy is,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{S}}_{surr}}\text{ = }\dfrac{-\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{sys}}}{{{\text{T}}_{\text{surr}}}}\text{ }$
Substitute the values in the equation. We have,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{S}}_{surr}}\text{ = }\dfrac{-\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{sys}}}{{{\text{T}}_{\text{surr}}}}\text{ }=\text{ }\dfrac{75\text{ }\times \text{1}{{\text{0}}^{\text{3}}}\text{ Jmo}{{\text{l}}^{-1}}\text{ }}{300\text{ K }}=\text{ }250\text{J}{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\text{ }$
Therefore from part A) and B). The entropy change associated with the formation of the product is equal to the sum of the entropy of the reaction and the entropy of the surrounding. Thus entropy is,
$\text{ }\Delta {{\text{S}}_{\text{rxn}}}\text{ = }\Delta {{\text{S}}^{\text{0}}}\text{ + }\Delta {{\text{S}}_{surr}}\text{ = }\left( -81\text{ + 250 } \right)\text{ }=\text{ 169 J }{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\text{ }$
Thus the entropy change associated with the reaction is equal to $\text{ 169 J }{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\text{ }$ .

Hence, (A) is the correct option.

Note: Not that, we have provided with the enthalpy of the formation of the reaction thus do not tempt to get the entropy directly from the $\text{ }\dfrac{\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{rxn}}}}{\text{T}}\text{ }$ . This condition is true only at the equilibrium condition when the Gibbs free energy of the reaction is zero. But remember that $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{H}}_{\text{rxn}}}$ is important to find out the entropy change associated with the surrounding.