C2.
A reversible reaction between hydrogen and iodine occurs as follows:
$$
\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \leftrightharpoons 2 \mathrm{HI}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{c}}=49.0 \text { at } 700 \mathrm{~K}
$$

Where $\mathrm{K}_{c}$ is the equilibrium constant.
(a) If a $10 \mathrm{~L}$ reaction vessel is fed with $1 \mathrm{~mole}$ of $\mathrm{H}_{2}$ and $1 \mathrm{~mole}_{2}$ at $700 \mathrm{~K}$, calculate the equilibrium concentrations of all components?
(b) If $\mathrm{k}_{\mathrm{r}}=1.20 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}$ at $700 \mathrm{~K}$, determine the value of $\mathrm{k}_{\mathrm{t}}$ at the same temperature? Where $\mathrm{k}_{\mathrm{f}}$ is the forward reaction rate constant and $\mathrm{k}_{r}$ is the reverse rate constant.
(c) If the forward reaction is endothermic, explain with justification how an increase in temperature effects the values of $\mathrm{K}_{\mathrm{c}}, \mathrm{k}_{\mathrm{f}}$ and $\mathrm{k}_{\mathrm{r}}$ ?

Question

C2.
A reversible reaction between hydrogen and iodine occurs as follows:
$$
\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \leftrightharpoons 2 \mathrm{HI}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{c}}=49.0 \text { at } 700 \mathrm{~K}
$$

Where $\mathrm{K}_{c}$ is the equilibrium constant.
(a) If a $10 \mathrm{~L}$ reaction vessel is fed with $1 \mathrm{~mole}$ of $\mathrm{H}_{2}$ and $1 \mathrm{~mole}_{2}$ at $700 \mathrm{~K}$, calculate the equilibrium concentrations of all components?
(b) If $\mathrm{k}_{\mathrm{r}}=1.20 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}$ at $700 \mathrm{~K}$, determine the value of $\mathrm{k}_{\mathrm{t}}$ at the same temperature? Where $\mathrm{k}_{\mathrm{f}}$ is the forward reaction rate constant and $\mathrm{k}_{r}$ is the reverse rate constant.
(c) If the forward reaction is endothermic, explain with justification how an increase in temperature effects the values of $\mathrm{K}_{\mathrm{c}}, \mathrm{k}_{\mathrm{f}}$ and $\mathrm{k}_{\mathrm{r}}$ ?

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Set up the equilibrium expression**: The equilibrium constant expression for the reaction $\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g)$ is given by $K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}$. ‖ ‖ 2 ⋮ **Calculate initial concentrations**: The initial concentrations of $\mathrm{H}_2$ and $\mathrm{I}_2$ are both $\frac{1 ext{ mole}}{10 ext{ L}} = 0.1 ext{ M}$. The initial concentration of $\mathrm{HI}$ is $0 ext{ M}$ since it has not yet formed. ‖ ‖ 3 ⋮ **Define change in concentrations**: Let $x$ be the change in concentration of $\mathrm{H}_2$ and $\mathrm{I}_2$ at equilibrium. Then the change in concentration of $\mathrm{HI}$ is $2x$. ‖ ‖ 4 ⋮ **Write the equilibrium concentrations**: At equilibrium, $[\mathrm{H}_2] = [\mathrm{I}_2] = 0.1 - x$ and $[\mathrm{HI}] = 2x$. ‖ ‖ 5 ⋮ **Apply the equilibrium constant**: Substitute the equilibrium concentrations into the equilibrium expression to get $49.0 = \frac{(2x)^2}{(0.1 - x)(0.1 - x)}$. ‖ ‖ 6 ⋮ **Solve for x**: Solve the quadratic equation to find the value of $x$. ‖ ‖ 7 ⋮ **Find equilibrium concentrations**: Use the value of $x$ to calculate the equilibrium concentrations of $\mathrm{H}_2$, $\mathrm{I}_2$, and $\mathrm{HI}$. ‖ ∻1 Answer∻ ⚹ [Insert equilibrium concentrations here] ⚹ ∻Key Concept∻ ⚹ Equilibrium concentrations are determined by applying the equilibrium constant to the changes in concentrations of reactants and products. ⚹ ∻Explanation∻ ⚹ The equilibrium constant expression relates the concentrations of reactants and products at equilibrium. By setting up a system of equations based on the stoichiometry of the reaction and the equilibrium constant, we can solve for the unknown concentrations. ⚹ ∻Solution∻ ‖ 1 ⋮ **Use the equilibrium constant expression**: The equilibrium constant $K_c$ is related to the rate constants of the forward ($k_f$) and reverse ($k_r$) reactions by the equation $K_c = \frac{k_f}{k_r}$. ‖ ‖ 2 ⋮ **Solve for the reverse rate constant**: Rearrange the equation to solve for $k_r$: $k_r = \frac{k_f}{K_c}$. ‖ ‖ 3 ⋮ **Substitute known values**: Substitute $k_f = 1.20 imes 10^{-3} ext{ M}^{-1} ext{s}^{-1}$ and $K_c = 49.0$ into the equation to find $k_r$. ‖ ∻2 Answer∻ ⚹ [Insert value of $k_r$ here] ⚹ ∻Key Concept∻ ⚹ The reverse rate constant can be calculated from the equilibrium constant and the forward rate constant. ⚹ ∻Explanation∻ ⚹ Knowing the equilibrium constant and the forward rate constant allows us to find the reverse rate constant by rearranging the relationship between these constants. ⚹ ∻Solution∻ ‖ 1 ⋮ **Understand Le Chatelier's Principle**: An increase in temperature for an endothermic reaction shifts the equilibrium to favor the formation of products, increasing $K_c$. ‖ ‖ 2 ⋮ **Predict the effect on rate constants**: According to the Arrhenius equation, an increase in temperature increases the rate constants $k_f$ and $k_r$, but the increase in $k_f$ will be more significant for an endothermic reaction. ‖ ∻3 Answer∻ ⚹ An increase in temperature will increase $K_c$, $k_f$, and $k_r$, with a more significant increase in $k_f$ for an endothermic reaction. ⚹ ∻Key Concept∻ ⚹ Temperature changes affect equilibrium constants and rate constants differently depending on the endothermic or exothermic nature of the reaction. ⚹ ∻Explanation∻ ⚹ For an endothermic reaction, increasing temperature shifts the equilibrium towards products, increasing $K_c$. The rate constants $k_f$ and $k_r$ also increase with temperature, but the forward rate constant increases more due to the reaction absorbing heat. ⚹◊What is the mathematical relationship between the forward rate constant ($k_f$) and the reverse rate constant ($k_r$) in a reversible chemical reaction?⍭ Generate me a similar question◊

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