Question

Write the balanced chemical equation for the reaction of sulphuric acid on each of the following-
(A) Potassium hydrogen carbonate
(B) Sulphur

Answer 1

Sulphuric acid reacts with Potassium hydrogen carbonate to give the products – water, carbon dioxide and potassium sulphate. Sulphuric acid reacts with sulphur to give the products – sulphur dioxide and water.

Complete step by step solution:
- (A) The reaction of sulphuric acid with Potassium hydrogen carbonate is an acid base neutralisation reaction. The sulphuric acid is the acid and potassium bicarbonate is the base. We get the products as water, carbon dioxide and potassium sulphate.
- As it is a very simple reaction, we can use the trial and error method.
- Let us first of all write the unbalanced chemical equation.
$KHC{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+C{{O}_{2}}+{{H}_{2}}O$
- We can see that there is one less potassium on the right side. So to balance the equation, we will put 2 as a coefficient there.
$2KHC{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+C{{O}_{2}}+{{H}_{2}}O$
- Now, we can see that there are 2 C atoms on the left and only 1 on the right. Moreover, the number of H atoms on the right is 4 whereas there are only 2 H atoms on the right. To balance this, we can put a coefficient of 2 on the carbon dioxide and a coefficient of 2 on the water molecule.
$2KHC{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2C{{O}_{2}}+2{{H}_{2}}O$
- We can see that the equation has been balanced.
(B) The reaction of sulphuric acid with sulphur is a redox reaction where both oxidation and reduction occurs. The S in 0 state gets oxidised to +2 state in sulphur dioxide and the S in +6 state in the sulphuric acid gets reduced to +2 state in sulphur dioxide.
$S+{{H}_{2}}S{{O}_{4}}\to S{{O}_{2}}+{{H}_{2}}O$
- For this reaction, the method used is called the half equation method. In this method, we write the reduction reaction and oxidation reaction occurring separately and then balance them. The reduction half and oxidation half can be written as,

& \text{Oxidation half : }{{\text{S}}^{0}}\to {{S}^{(4+)}}{{O}_{2}} \\\ & \operatorname{Re}\text{duction half : }{{\text{H}}_{2}}{{S}^{(6+)}}{{O}_{4}}\to {{S}^{(4+)}}{{O}_{2}} \\\ \end{aligned}$$

- After writing the oxidation and reduction halves, we have to balance the number of the oxidised and reduced elements in the half equations. For this question the number of S atoms is already equal in both half equations.
- The next step is to balance the elements other than O and H. In this question there are no such elements.
- Next, we have to balance the number of oxygen atoms. For doing this, we add ${{H}_{2}}O$ to the side lacking oxygen atoms.

& \text{Oxidation half : }{{\text{S}}^{0}}+2{{H}_{2}}O\to {{S}^{(4+)}}{{O}_{2}} \\\ & \operatorname{Re}\text{duction half : }{{\text{H}}_{2}}{{S}^{(6+)}}{{O}_{4}}\to {{S}^{(4+)}}{{O}_{2}}+2{{H}_{2}}O \\\ & \\\ \end{aligned}$$ \- Next, we have to balance the number of H atoms in both half equations. As it is an acidic medium, we add ${{H}^{+}}$ ions to the side lacking H atoms. $$\begin{aligned} & \text{Oxidation half : S}+2{{H}_{2}}O\to S{{O}_{2}}+4{{H}^{+}} \\\ & \operatorname{Re}\text{duction half : }{{\text{H}}_{2}}S{{O}_{4}}+2{{H}^{+}}\to S{{O}_{2}}+2{{H}_{2}}O \\\ & \\\ \end{aligned}$$ \- Next, we have to balance the charge. This is done simply by adding electrons in the required amount to the side having positive charge. $$\begin{aligned} & \text{Oxidation half : S}+2{{H}_{2}}O\to S{{O}_{2}}+4{{H}^{+}}+4{{e}^{-}} \\\ & \operatorname{Re}\text{duction half : }{{\text{H}}_{2}}S{{O}_{4}}+2{{H}^{+}}+2{{e}^{-}}\to S{{O}_{2}}+2{{H}_{2}}O \\\ \end{aligned}$$ \- Now, we have to make the number of electrons equal in both half equations and then add them. In this case we multiply the reduction equation by 2 and then add it to the oxidation equation. This gives us $$\text{S}+2{{H}_{2}}O+2{{\text{H}}_{2}}S{{O}_{4}}+4{{H}^{+}}+4{{e}^{-}}\to 3S{{O}_{2}}+4{{H}^{+}}+4{{e}^{-}}+4{{H}_{2}}O$$ \- Now, we can cancel out the common terms to get the final balanced equation. $$\text{S}+2{{\text{H}}_{2}}S{{O}_{4}}\to 3S{{O}_{2}}+2{{H}_{2}}O$$ **Note:** For redox reactions, the half equation method is used as the trial and error method is hard in case of redox reactions. You can also balance the redox reaction by using trial and error methods if you are able to.