Q105CP - Chemistry: Molecular Nature Of Matter And Change

Question

The molecular scene depicts a gaseous equilibrium mixture at$\mathbf{460}\mathbf{°}\mathit{C}$  for the reaction of${\mathit{H}}_{\mathbf{2}}$  (blue) and  ${\mathit{I}}_{\mathbf{2}}$(purple) to form HI. Each molecule represents  $\mathbf{0}\mathbf{.}\mathbf{010}$mol and the container volume is $\mathbf{1}\mathbf{.}\mathbf{0}$L. 

(a) Is${\mathit{K}}_{\mathbf{c}}\mathbf{>}\mathbf{,}\mathbf{=}\mathbf{,}\mathbf{ }\mathit{o}\mathit{r}\mathbf{<}\mathbf{1}$ ? 

(b) Is${\mathit{K}}_{\mathbf{p}}\mathbf{>}\mathbf{,}\mathbf{=}\mathbf{,}\mathit{o}\mathit{r}\mathbf{<}{\mathit{K}}_{\mathbf{c}}$ ? 

(c) Calculate$\mathbf{∆}{\mathit{G}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}$ . 

(d) How would the value of $\mathbf{∆}{\mathit{G}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}$ change if the purple molecules represented ${\mathit{H}}_{\mathbf{2}}$and the blue${\mathit{I}}_{\mathbf{2}}$ ? Explain.

Step-by-Step Solution

Answer

a) Because the concentration equilibrium constant equals$50$ , the reaction is product-favorable, and $K_c>1$.

b) For the specified reaction and conditions, $K_p=K_c$.

c) $\text{ - 16.78kJ}$is the$∆G_{rxn}^{°}$ .

d) Gibb's energy is unaffected since the equilibrium constant remains unchanged.

Step 1: Definition of Equilibrium

Equilibrium refers to a situation in which the concentrations of reactants and products do not change significantly. Although there appears to be no change in equilibrium, this does not imply that all chemical reactions have stopped.

Step 2: Determining

(a)The response that was given:

${\text{H}}_{\text{2}}{\text{(g) + I}}_{\text{2}}\text{(g)}\to \text{2HI(g)}$

• Calculate the $K_c$ ratio of product to reactant concentrations, which is the equilibrium constant (concentration). Calculating $K_c$for a specific response

${\text{K}}_{\text{c}}\text{ = }\frac{{\text{[HI]}}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]}$

• Each component's concentration can be found as

$\text{c = }\frac{\text{n}}{\text{V}}$

• As a result, if 1 molecule represents $0.010$mol of substance in $1$L total volume, concentration for each component can be determined based on the reaction.

- For the $H_2$(blue),

$$\begin{gathered} \left[H_2\right]=\frac{1\times 0.010mol}{1L} \\ \left[H_2\right]=0.010M \end{gathered}$$

- For the$I_2$ (purple),

$$\begin{gathered} \left[I_2\right]=\frac{2\times 0.010mol}{1L} \\ \left[I_2\right]=0.020M \end{gathered}$$

Step 3: Calculating K c &#62; , = , o r &#60; 1

- For the $HI$ (blue-purple),

$$\begin{gathered} \left[HI\right]=\frac{10\times 0.010mol}{1L} \\ \left[HI\right]=0.10M \end{gathered}$$

• Last but not least, $K_c$can be determined.

$$\begin{gathered} {\text{K}}_{\text{c}}\text{ = }\frac{{\text{[HI]}}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]}\text{=}\frac{{\text{[0.10M]}}^{\text{2}}}{\text{[0.010M][0.020M]}} \\ {K}_c=50 \end{gathered}$$

• The product-favorable reaction has a concentration equilibrium constant of $50$, indicating that the reaction is product-favorable.

$K_c>1$

Step 4: Determining K p &#62; , = , o r &#60; K c

(b) For partial pressures, the$K_p$ is a so-called equilibrium constant. Because we have a gaseous mixture equilibrium with all components in a gaseous state and a volume of$1L$ , $K_p$equals$K_c$ .

${\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}\text{ = 50}$

Step 5: Determining &#8710; G r x n &#176;

(c) Calculate the reaction's Gibbs change under standard conditions ($298K$),

$$\begin{gathered} ∆G_{rxn}^{°}=\sum _{}^{}∆G_{products}^{°}-\sum _{}^{}∆G_{reac\tan ts}^{°} \\ ∆G_{rxn}^{°}=\left[2mol\times ∆G_{H{I}_{\left(g\right)}}^{°}\right]-\left[1mol\times ∆G_{H_{2_{\left(g\right)}}}^{°}+1mol\times ∆G_{I_{2_{\left(g\right)}}}^{°}\right] \end{gathered}$$

• Since hydrogen gas is in a standard condition, use Appendix B.

$$\begin{gathered} ∆G_{rxn}^{°}=\left[2mol\times \left(1.30kJ/mol\right)\right]-\left[1mol\times \left(0.00kJ/mol\right)+1mol\times \left(19.38kJ/mol\right)\right] \\ ∆G_{rxn}^{°}=-16.78kJ \end{gathered}$$

$\text{ - 16.78kJ}$ is the $∆G_{rxn}^{°}$.

Step 6: Determining how the value of &#8710; G r x n &#176; change

d) The reaction equilibrium constant K would not change if the colors were changed to purple (for$H_2$ ) and blue (for $I_2$):

As a result, the component concentrations would be altered.

- For the$H_2$ (purple),

$$\begin{gathered} \left[H_2\right]=\frac{2\times 0.010mol}{1L} \\ \left[H_2\right]=0.020M \end{gathered}$$

- For the$I_2$ (blue),

$$\begin{gathered} \left[I_2\right]=\frac{1\times 0.010mol}{1L} \\ \left[I_2\right]=0.010M \end{gathered}$$

• The concentration of HI (blue-purple) remains constant.
• Last but not least, $K_{c_2}$can be recalculated.

$$\begin{gathered} {\text{K}}_{\text{c,2}}\text{ = }\frac{{\text{[HI]}}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]}\text{=}\frac{{\text{[0.10M]}}^{\text{2}}}{\text{[0.020M][0.010M]}} \\ {\text{K}}_{\text{c,2}}\text{ = 50} \end{gathered}$$

• As previously stated, $K_c$the would not alter. As a result, the Gibb's energy would be unaffected.