**For this problem, we need to solve for **ΔS°rxn,** **ΔG°rxn, and determine which direction the reaction is spontaneous.

For the reaction at** T = 25.0°C: **

**3C _{2}H_{2}(g) → C_{6}H_{6}(I)**

The** ΔH°rxn = -633.1 kJ•mol ^{-1 }(or kJ/mol) **

Recall that the equation for the **standard entropy, ΔS°rxn **is**:**

$\overline{){\mathbf{\u2206}}{\mathit{S}}{{\mathbf{\xb0}}}_{\mathbf{r}\mathbf{x}\mathbf{n}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathit{S}}{{\mathbf{\xb0}}}_{{\mathbf{products}}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}{\mathit{S}}{{\mathbf{\xb0}}}_{{\mathbf{reactants}}}}$

The **standard free energy change of a reaction (ΔG˚ _{rxn})** is given by the following equation:

$\overline{){\mathbf{\u2206}}{\mathit{G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{\u2206}}{\mathbf{H}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{-}}{\mathbf{T}}{\mathbf{\u2206}}{\mathbf{S}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}}$

From the given table, we can find the **ΔS˚** of each reactant and product and use these values to calculate **ΔS˚ _{rxn}**

Calculate the standard entropy, ΔS°_{rxn}, of the following reaction at 25.0°C using the data in this table. The standard enthalpy of the reaction, ΔH°_{rxn}, is -633.1 kJ•mol^{-1}.

3C_{2}H_{2}(g) → C_{6}H_{6}(I)

ΔS°_{rxn =}

Then, calculate the standard Gibbs free energy of the reaction, ΔG°rxn.

ΔG°_{rxn = }

Finally, determine which direction the reaction is spontaneous as written at 25.0°C and standard pressure.

a. forward

b. reverse

c. both

d. neither

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