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Calculate the free energy change for the same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?

Short Answer

Expert verified

The calculated boiling point of CS2is \(323K\).

Step by step solution

01

Define the enthalpy of the reaction

The change in Gibbs free energy is as follows:

\(\Delta {\rm{G}} = \Delta {\rm{H}} - {\rm{T}}\Delta {\rm{S}}\)

where,

\({\bf{\Delta G }}\)= change in Gibbs free energy,

\({\bf{\Delta H}}\)= change in enthalpy,

T = absolute temperature in Kelvin, and

\({\bf{\Delta S}}\)= change in entropy.

The Gibbs free energy change is used to determine the spontaneity of a process. It is expressed in terms of the enthalpy and the entropy of a system.

Entropy is the degree of disorderness or randomness in a given system. The entropy change during a transition phase is expressed as

\(\Delta S = \frac{{\Delta H}}{T}\)

02

Determine the estimate the boiling pont of CS2

Given:

\({\rm{\Delta }}{{\rm{H}}^ \circ }f({\rm{KJ/mol}})\)

\({{\rm{S}}^ \circ }f({\rm{J/Kmol}})\)

\({\rm{CS2}}\)(g)

115.3

237.8

\({\rm{CS2}}\)(l)

87.3

151

Carbon disulfide vaporization

\(C{S_{2(l)}} \to C{S_{2(g)}}\)

Solving for enthalpy change of vaporization:

\(\Delta {H_{vap}} = \Delta H_{f\left( {C{S_{2(g)}}} \right.}^o - \Delta H_{f\left( {C{S_{2(l)}}} \right.}^o = 115.3 - 87.3 = 28\frac{{kJ}}{{{\rm{ mole }}}}\)

Solving for entropy change of vaporization:

\(\Delta {S_{vap}} = \Delta S_{f\left( {C{S_{2(g)}}} \right.}^o - \Delta S_{f\left( {C{S_{2(l)}}} \right.}^o = 237.8 - 151 = 86.8\frac{J}{{{\rm{ mole }} \cdot K}}\)

Solving for boiling point using the entropy definition for phase changes:

\(\Delta {S_{vap}} = \frac{{\Delta {H_{vap}}}}{{{T_b}}}\)

\(\begin{array}{l}{T_b} = \frac{{\Delta {H_{vap}}}}{{\Delta {S_{vap}}}}\\ = \frac{{28000\frac{J}{{{\rm{ mole }}}}}}{{86.8\frac{J}{{{\rm{ mole }} \cdot K}}}}\\ = 323K\end{array}\).

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Most popular questions from this chapter

Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.

(a) \({{\bf{N}}_{\bf{2}}}{\bf{(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}}\)

(b) \({\bf{HCl(g) + N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}} \to {\bf{N}}{{\bf{H}}_{\bf{4}}}{\bf{Cl(s)}}\)

(c) \({\left( {{\bf{N}}{{\bf{H}}_{\bf{4}}}} \right)_{\bf{2}}}{\bf{C}}{{\bf{r}}_{\bf{2}}}{{\bf{O}}_{\bf{7}}}{\bf{(s)}} \to {\bf{C}}{{\bf{r}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 4}}{{\bf{H}}_{\bf{2}}}{\bf{O(g) + }}{{\bf{N}}_{\bf{2}}}{\bf{(g)}}\)

(d) \({\bf{2Fe(s) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}}\)

Consider the system shown in Figure 16.9. What is the change in entropy for the process where all the energy is transferred from the hot object (AB) to the cold object (CD)

What is a spontaneous reaction?

Calculate the standard entropy change for the following process:

\({{\bf{H}}_{\bf{2}}}{\bf{(g) + }}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{6}}}{\bf{(g)}}\)

Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.

(a) Ammonia\({\bf{:}}{\rm{ }}{\bf{2N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(b) Diborane\({\bf{:}}{\rm{ }}{{\bf{B}}_{\bf{2}}}{{\bf{H}}_{\bf{6}}}{\bf{(g)}} \to {\bf{2\;B(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(c) Hydrazine: \({{\bf{N}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(d) Hydrogen peroxide: \({{\bf{H}}_{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(l)}} \to {{\bf{H}}_{\bf{2}}}{\bf{O(g) + }}\frac{{\bf{1}}}{{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

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