2) Consider the following:
$$
2 \mathrm{HI}_{(\mathrm{g})} \rightleftarrows \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \quad \mathrm{K}_{e q}=81.0
$$

A $1.0 \mathrm{~L}$ flask is initially filled with $2.00 \mathrm{~mol}_{2}$ and $2.00 \mathrm{~mol}_{2}$. Calculate the $\left[\mathrm{H}_{2}\right]$ at equilibrium. (5 marks)
3) Which way will the equilibrium shift (if any) when the following changes are made:
$$
4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})}+\text { heat } \rightleftarrows 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}
$$
a) $[\mathrm{NO}]$ is decreased (1 mark)
b) Temperature is decreased (1 mark)
c) The volume of the container is increased (1 mark)
d) A catalyst is added (1 mark)

Question

2) Consider the following:
$$
2 \mathrm{HI}_{(\mathrm{g})} \rightleftarrows \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \quad \mathrm{K}_{e q}=81.0
$$

A $1.0 \mathrm{~L}$ flask is initially filled with $2.00 \mathrm{~mol}_{2}$ and $2.00 \mathrm{~mol}_{2}$. Calculate the $\left[\mathrm{H}_{2}\right]$ at equilibrium. (5 marks)
3) Which way will the equilibrium shift (if any) when the following changes are made:
$$
4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})}+\text { heat } \rightleftarrows 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}
$$
a) $[\mathrm{NO}]$ is decreased (1 mark)
b) Temperature is decreased (1 mark)
c) The volume of the container is increased (1 mark)
d) A catalyst is added (1 mark)

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equation**: The balanced chemical equation for the dissociation of hydrogen iodide is $2 \mathrm{HI(g)} \rightleftarrows \mathrm{H}_{2(g)} + \mathrm{I}_{2(g)}$. ‖ ‖ 2 ⋮ **Set up the ICE table**: To calculate the equilibrium concentration of $\mathrm{H}_{2}$, we use an ICE (Initial, Change, Equilibrium) table. Initial concentrations are $[\mathrm{H}_{2}] = 0$ and $[\mathrm{I}_{2}] = 0$ since we start with only HI. The change will be $-x$ for HI and $+x$ for both $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$, as they are produced. ‖ ‖ 3 ⋮ **Write the expression for $K_c$**: The equilibrium constant expression is $K_c = \frac{[\mathrm{H}_{2}][\mathrm{I}_{2}]}{[\mathrm{HI}]^2}$. ‖ ‖ 4 ⋮ **Calculate the change in concentration, $x$**: Plugging the equilibrium concentrations into the $K_c$ expression gives $81.0 = \frac{x^2}{(2.00 - x)^2}$. Solving for $x$ gives the change in concentration of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ at equilibrium. ‖ ‖ 5 ⋮ **Solve for $x$ and find $[\mathrm{H}_{2}]$**: Solving the equation from step 4 for $x$ will give us the equilibrium concentration of $\mathrm{H}_{2}$, which is $x$. ‖ ∻2 Answer∻ ⚹ The equilibrium concentration of $\mathrm{H}_{2}$ is $x$ mol/L, where $x$ is the solution to the equation $81.0 = \frac{x^2}{(2.00 - x)^2}$. ⚹ ∻Key Concept∻ ⚹ Equilibrium concentration calculations involve setting up an ICE table and solving for the change in concentration using the equilibrium constant expression. ⚹ ∻Explanation∻ ⚹ The equilibrium constant expression relates the concentrations of reactants and products at equilibrium. By setting up an ICE table and solving the $K_c$ expression, we can find the equilibrium concentrations. ⚹ ∻Solution∻ ‖ 1 ⋮ **Analyze the effect of decreasing $[\mathrm{NO}]$**: According to Le Chatelier's principle, if the concentration of a product is decreased, the equilibrium will shift to the right to produce more product. ‖ ∻3a Answer∻ ⚹ The equilibrium will shift to the right, producing more $\mathrm{NO}$ and $\mathrm{H}_{2}O$. ⚹ ‖ 2 ⋮ **Analyze the effect of decreasing temperature**: Since the forward reaction is exothermic (releases heat), decreasing the temperature will shift the equilibrium to the right to produce more heat. ‖ ∻3b Answer∻ ⚹ The equilibrium will shift to the right, favoring the formation of $\mathrm{NO}$ and $\mathrm{H}_{2}O$. ⚹ ‖ 3 ⋮ **Analyze the effect of increasing the volume**: Increasing the volume of the container will decrease the pressure, shifting the equilibrium towards the side with more moles of gas. ‖ ∻3c Answer∻ ⚹ The equilibrium will shift to the left, favoring the formation of $\mathrm{NH}_{3}$ and $\mathrm{O}_{2}$. ⚹ ‖ 4 ⋮ **Analyze the effect of adding a catalyst**: A catalyst lowers the activation energy for both the forward and reverse reactions, but it does not shift the position of equilibrium. ‖ ∻3d Answer∻ ⚹ The addition of a catalyst will not shift the equilibrium; it will only increase the rate at which equilibrium is achieved. ⚹ ∻Key Concept∻ ⚹ Le Chatelier's principle predicts the direction of the shift in equilibrium when a system is subjected to a change in concentration, temperature, pressure, or the addition of a catalyst. ⚹ ∻Explanation∻ ⚹ Changes in concentration, temperature, and pressure can shift the equilibrium to the right or left, while a catalyst only affects the rate of the reaction, not the position of equilibrium. ⚹◊What is the expression for the equilibrium constant (Kc) in the given reaction involving hydrogen iodide?⍭ Generate me a similar question◊

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