Question

What property of Sulphuric acid is shown by the reaction of concentrated Sulphuric acid when heated with

1. Potassium nitrate
2. Carbon.

Answer 1

Part 1: Concentrated Sulphuric acid:

1. It can act as dehydrating agent, which means concentrated Sulphuric acid can remove the hydrogen and oxygen atom (which forms water molecule) from a compound.
2. It can act as an oxidizing agent, concentrated Sulphuric acid oxidizes the metal to form Sulphur dioxide gas.
3. It is also non-volatile, as it displaces high volatile acid from its salt solution.

Part 2: Potassium nitrate reacts with conc. Sulphuric acid:

1. $$\begin{gathered} {\mathrm{KNO}}_3\left(\mathrm{aq}\right)+\mathrm{conc}.{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{KHSO}}_4\left(\mathrm{s}\right)+{\mathrm{HNO}}_3\left(\mathrm{aq}\right) \\ \mathrm{Potassium}\mathrm{nitrate}+\mathrm{conc}.\mathrm{Sulphuric}\mathrm{acid}\to \mathrm{Potassium}\mathrm{bisulphate}+\mathrm{Nitric}\mathrm{acid} \end{gathered}$$
2. Concentrated Sulphuric acid displaced ${\mathrm{HNO}}_3$ which is highly volatile from the salt of ${\mathrm{KNO}}_3$.
3. This shows the non-volatility of concentrated Sulphuric acid.

Part 3: Carbon reacts with concentrated Sulphuric acid:

1. $$\begin{gathered} \mathrm{C}\left(\mathrm{s}\right)+\mathrm{conc}.{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \mathrm{Carbon}+\mathrm{conc}.\mathrm{Sulphuric}\mathrm{acid}\to \mathrm{Carbon}\mathrm{dioxide}+\mathrm{Sulphur}\mathrm{dioxide}+\mathrm{Water} \end{gathered}$$
2. Here concentrated sulphuric acid oxidises $\mathrm{C}$ to ${\mathrm{CO}}_2$ .
3. Thus the given reaction shows oxidising nature.