Explain the effect of the following on the equilibrium: N2(g) + 3H2(g) <=> 2NH3(g) (ΔH= -92Kj/mol). i. Increase in pressure:

Here are the energy change diagrams for the forward and backward reactions:

Forward reaction:

$$\ce{NH4Cl(s) -> NH3(g) + HCl(g)}$$

This is an endothermic reaction, meaning it absorbs energy. The products have higher potential energy than the reactants.

\begin{tikzpicture}[scale=0.8] % Axes \draw[->] (0,0) -- (6,0) node[right] {Reaction Progress}; \draw[->] (0,0) -- (0,5) node[above] {Potential Energy}; % Reactants \draw[thick] (0.5,1) -- (1.5,1) node[left,yshift=-0.5cm] {$\ce{NH4Cl(s)}$}; % Products \draw[thick] (4.5,4) -- (5.5,4) node[right,yshift=0.5cm] {$\ce{NH3(g) + HCl(g)}$}; % Reaction path \draw[thick] (1.5,1) .. controls (2.5,4.5) and (3.5,4.5) .. (4.5,4); % Activation Energy (Ea) \draw[<->] (1.7,1) -- (1.7,4.5); \node at (2.2,2.75) {$\mathrm{E_{a,f}}$}; % Heat of Reaction (ΔH) \draw[<->] (5.7,1) -- (5.7,4); \node at (6.2,2.5) {$\Delta H_f$}; \end{tikzpicture}

Backward reaction:

$$\ce{NH3(g) + HCl(g) -> NH4Cl(s)}$$

This is an exothermic reaction, meaning it releases energy. The products have lower potential energy than the reactants.

\begin{tikzpicture}[scale=0.8] % Axes \draw[->] (0,0) -- (6,0) node[right] {Reaction Progress}; \draw[->] (0,0) -- (0,5) node[above] {Potential Energy}; % Reactants \draw[thick] (0.5,4) -- (1.5,4) node[left,yshift=-0.5cm] {$\ce{NH3(g) + HCl(g)}$}; % Products \draw[thick] (4.5,1) -- (5.5,1) node[right,yshift=0.5cm] {$\ce{NH4Cl(s)}$}; % Reaction path \draw[thick] (1.5,4) .. controls (2.5,4.5) and (3.5,4.5) .. (4.5,1); % Activation Energy (Ea) \draw[<->] (1.7,4) -- (1.7,4.5); \node at (2.2,4.25) {$\mathrm{E_{a,r}}$}; % Heat of Reaction (ΔH) \draw[<->] (5.7,1) -- (5.7,4); \node at (6.2,2.5) {$\Delta H_r$}; \end{tikzpicture}