The melting point of $\text{Al}_2\text{O}_3$ is $2072^\circ\text{C}$. The melting point of $\text{P}_4\text{O}_{10}$ is $340^\circ\text{C}$. Explain why these two compounds have different melting points.
A $5.00\,\text{dm}^3$ sealed flask contains $0.400\,\text{mol}$ of CO(g) and $0.800\,\text{mol}$ of $\text{H}_2$(g) together with an $\text{Al}_2\text{O}_3$ catalyst. The flask is heated to a temperature of $290^\circ\text{C}$ and left until equilibrium is reached. Equation 1 shows the reaction. $\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}$ The equilibrium constant, $K_c$, for equation 1 is given. $K_c = \dfrac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}$ State the units of $K_c$.
At equilibrium, the mixture contains $0.280\,\text{mol}$ of $\text{CH}_3\text{OH(g)}$. Calculate the value of $K_c$. Give your answer to three significant figures.
State and explain what happens, if anything, to the value of $K_c$ when the overall pressure in the sealed flask is increased.
$\text{P}_4\text{O}_{10}$ catalyses the reversible reaction of CO with $\text{H}_2$ to form $\text{CH}_3\text{OH}$. Equation 1: $\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}$ $\text{P}_4\text{O}_{10}$ then acts as a dehydrating agent, causing $\text{CH}_3\text{OH}$ to form $\text{CH}_3\text{OCH}_3$. Explain how the presence of a catalyst affects a chemical reaction.
Construct an equation showing the dehydration of $\text{CH}_3\text{OH}$ to produce $\text{CH}_3\text{OCH}_3$.
Write an equation showing the reaction of $\text{P}_4\text{O}_{10}$ with an excess of water.