6)

A reaction between hydrogen and iodine occurs as follows:
$$
\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})=2 \mathrm{HI}(\mathrm{g}) \quad K_{c}=57.0 \text { at } 700 \mathrm{~K}
$$

If a 10 litre reaction vessel is fed with $1 \mathrm{~mole}$ of $\mathrm{H}_{2}$ and $1 \mathrm{~mole}_{2}$ at $700 \mathrm{~K}$, what are the equilibrium concentrations of all components?
$$
\left[c_{\mathrm{H} 2}=c_{12}=0.021 \mathrm{M} ; c_{\mathrm{HI}}=0.158 \mathrm{M}\right]
$$

Question

6)

A reaction between hydrogen and iodine occurs as follows:
$$
\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})=2 \mathrm{HI}(\mathrm{g}) \quad K_{c}=57.0 \text { at } 700 \mathrm{~K}
$$

If a 10 litre reaction vessel is fed with $1 \mathrm{~mole}$ of $\mathrm{H}_{2}$ and $1 \mathrm{~mole}_{2}$ at $700 \mathrm{~K}$, what are the equilibrium concentrations of all components?
$$
\left[c_{\mathrm{H} 2}=c_{12}=0.021 \mathrm{M} ; c_{\mathrm{HI}}=0.158 \mathrm{M}\right]
$$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equation**: The balanced chemical equation for the reaction between hydrogen gas and iodine gas to form hydrogen iodide is $$ \mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)} $$. ‖ ‖ 2 ⋮ **Set up the ICE table**: To find the equilibrium concentrations, we use an ICE (Initial, Change, Equilibrium) table. Initially, we have 1 mole of $\mathrm{H_2}$ and 1 mole of $\mathrm{I_2}$ in a 10-liter vessel, so their concentrations are both $0.1 \, \mathrm{M}$. The initial concentration of $\mathrm{HI}$ is $0 \, \mathrm{M}$. At equilibrium, the concentration of $\mathrm{H_2}$ and $\mathrm{I_2}$ will have decreased by $x$ moles/L, and the concentration of $\mathrm{HI}$ will have increased by $2x$ moles/L. ‖ ‖ 3 ⋮ **Apply the equilibrium constant expression**: The equilibrium constant expression for the reaction is $$ K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} $$. Given that $ K_c = 57.0 $ at $ 700 \, \mathrm{K} $, we can set up the equation $ 57.0 = \frac{(2x)^2}{(0.1 - x)(0.1 - x)} $. ‖ ‖ 4 ⋮ **Solve for $ x $**: Solving the quadratic equation for $ x $ will give us the change in concentration for $\mathrm{H_2}$ and $\mathrm{I_2}$ and the formation of $\mathrm{HI}$ at equilibrium. ‖ ‖ 5 ⋮ **Calculate equilibrium concentrations**: Once $ x $ is found, we can calculate the equilibrium concentrations: $$ [\mathrm{H_2}] = [\mathrm{I_2}] = 0.1 - x $$ and $$ [\mathrm{HI}] = 2x $$. Given the provided answer, $ x = 0.079 \, \mathrm{M} $, so the equilibrium concentrations are $ [\mathrm{H_2}] = [\mathrm{I_2}] = 0.021 \, \mathrm{M} $ and $ [\mathrm{HI}] = 0.158 \, \mathrm{M} $. ‖ ∻Answer∻ ⚹ [H₂] = [I₂] = 0.021 M; [HI] = 0.158 M ⚹ ∻Key Concept∻ ⚹ Equilibrium concentrations are determined using an ICE table and the equilibrium constant expression. ⚹ ∻Explanation∻ ⚹ The equilibrium concentrations are calculated by setting up an ICE table, applying the equilibrium constant expression, and solving for the change in concentration (x), which is then used to find the final concentrations of all species at equilibrium. ⚹◊What is the expression for the equilibrium constant $K_c$ in terms of the concentrations of reactants and products in a chemical reaction?⍭ Generate me a similar question◊

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