H_{2}SO_{4} (Sulphuric acid) is the strong acid and Fe_{2}S_{3}(Ferric sulphide) is an unstable artificial material. Let us know some facts about the H_{2}SO_{4} + Fe_{2}S_{3}

**Sulphuric acid is also known as oil of vitriol. It is colorless, odorless viscous liquid. Ferric sulphide composed with the elements Sulphur & Iron, it is also known as Iron(III) sulphide. It is solid, black powder.**

In this article we will discuss about the reaction between Sulphuric acid and Ferric sulphide, balancing of equation, titration & ionic equation of the reaction.

**What is the product of H**_{2}SO_{4} & Fe_{2}S_{3}

_{2}SO

_{4}& Fe

_{2}S

_{3}

**Ferric sulphate and Hydrogen sulphide are formed when Sulphuric acid reacts with Ferric sulphide.**

**3H _{2}SO_{4} + Fe_{2}S_{3} = Fe_{2}(SO_{4})_{3} + 3H_{2}S**

**What type of reaction is H**_{2}SO_{4} + Fe_{2}S_{3}

_{2}SO

_{4}+ Fe

_{2}S

_{3}

**H _{2}SO_{4} + Fe_{2}S_{3} is a displacement reaction.**

**How to balance H**_{2}SO_{4} + Fe_{2}S_{3}

_{2}SO

_{4}+ Fe

_{2}S

_{3}

**The reaction is balanced using following steps**

**H _{2}SO_{4} + Fe_{2}S_{3} = Fe_{2}(SO_{4})_{3} + H_{2}S**

**The number of atoms present on both sides is counted, which should be the same.**

Reactant Side | Product Side |
---|---|

2-Hydrogen atom | 2-Hydrogen atom |

4-Oxygen atom | 12-Oxygen atom |

4-Sulphur atom | 4-Sulphur atom |

2-Iron atom | 2-Iron atom |

**Number of atoms present in reactant side and product side**

**The number of Oxygen atoms of both the sides of the reaction balance by adding coefficient 3 to the Sulphuric acid molecule at the reactant side.****3H**_{2}SO_{4}+ Fe_{2}S_{3}= Fe_{2}(SO_{4})_{3}+ H_{2}S**The number of Hydrogen atoms of both sides of the reaction balance by adding coefficient 3 to the Hydrogen sulphide molecule at product side**.**3H**_{2}SO_{4}+ Fe_{2}S_{3}= Fe_{2}(SO_{4})_{3}+ 3H_{2}S**So the balanced chemical reaction is****3H**_{2}SO_{4}+ Fe_{2}S_{3}= Fe_{2}(SO_{4})_{3}+ 3H_{2}S

**H**_{2}SO_{4} + Fe_{2}S_{3 }Titration

_{2}SO

_{4}+ Fe

_{2}S

_{3 }Titration

**H _{2}SO_{4} + Fe_{2}S_{3} do not undergo titration because there is no equivalence point.**

**H**_{2}SO_{4} + Fe_{2}S_{3} net ionic equation

_{2}SO

_{4}+ Fe

_{2}S

_{3}net ionic equation

**The net ionic equation**

**3H _{2}^{+}(aq) + Fe2S_{3} (s) = 2Fe^{3+} (aq) + 3H_{2}S (g)**

**The net ionic equation is derived using the steps mentioned below.**

**The balanced equation with phases for H**_{2}SO_{4}+ Fe_{2}S_{3}

**3H**_{2}SO_{4}(aq) + Fe_{2}S_{3}(s) = Fe_{2}(SO_{4})_{3}(aq) + 3H_{2}S (g)

**Then the equation written in splitting of compounds**

**3H**_{2}^{+}(aq) + 3SO_{4}^{2-}(aq) + Fe_{2}S_{3}(s) = 2Fe^{3+}(aq) + 3SO_{4}^{2-}(aq) + 3H_{2}S (g)

**The net ionic equation is as follows by removing the spectator ions (ions those are same on reactant & product sides)**

**3H**_{2}^{+}(aq) + Fe2S_{3}(s) = 2Fe^{3+}(aq) + 3H_{2}S (g)

**H**_{2}SO_{4} + Fe_{2}S_{3 }conjugate pairs

_{2}SO

_{4}+ Fe

_{2}S

_{3 }conjugate pairs

**The conjugate base of H**_{2}SO_{4}is SO_{4}^{2-}**The conjugate base of Fe**^{2}S^{3}is S_{3}^{2-}

**H**_{2}SO_{4} & Fe_{2}S_{3} Intermolecular forces

_{2}SO

_{4}& Fe

_{2}S

_{3}Intermolecular forces

**Hydogen bonding, dipole-dipole interaction and Van der Waals dispersion are the intermolecular forces in H**_{2}SO_{4}**Ferric sulphide is covalently bonded to each other**

**H**_{2}SO_{4} + Fe_{2}S_{3} reaction enthalpy

_{2}SO

_{4}+ Fe

_{2}S

_{3}reaction enthalpy

**The reaction enthalpy of H _{2}SO_{4} + Fe_{2}S_{3} is -103.33**

**KJ/mol.**

Compound | Moles | Enthalpy of formation, ΔH^{0}_{f} (KJ/mol) |
---|---|---|

H_{2}SO_{4} | 3 | -814 |

Fe_{2}S_{3} | 1 | -101.67 |

Fe_{2}(SO_{4})_{3} | 1 | -2585.2 |

H_{2}S | 3 | -20.6 |

**Bond enthalpy values**

**The standard enthalpy of a reaction is calculated using the following formula-****ΔH**^{0}_{f (reaction)}= ΣΔH^{0}_{f (product)}– ΣΔH^{0}_{f (reactants)}**Thus, Enthalpy change = -103.33****KJ/mol**

**Is** **H**_{2}SO_{4} + Fe_{2}S_{3} a buffer solution

_{2}SO

_{4}+ Fe

_{2}S

_{3}a buffer solution

**H _{2}SO_{4} + Fe_{2}S_{3} reaction will not form a buffer solution because Sulphuric acid is a strong acid and there can only be present a weak acid or base in buffer solution.**

**Is** **H**_{2}SO_{4} + Fe_{2}S_{3} a complete reaction

_{2}SO

_{4}+ Fe

_{2}S

_{3}a complete reaction

**H _{2}SO_{4} + Fe_{2}S_{3} is a complete reaction as products like Ferric sulphate and Hydrogen sulphide are formed.**

**Is** **H**_{2}SO_{4} + Fe_{2}S_{3} an exothermic or endothermic reaction

_{2}SO

_{4}+ Fe

_{2}S

_{3}an exothermic or endothermic reaction

**H _{2}SO_{4 }+ Fe_{2}S_{3}**

**is an exothermic reaction because it releases heat to the surrounding during reaction.**

**Is** **H**_{2}SO_{4} + Fe_{2}S_{3} a redox reaction

_{2}SO

_{4}+ Fe

_{2}S

_{3}a redox reaction

**H _{2}SO_{4 }+ Fe_{2}S_{3} is not a redox reaction because there is no change in oxidation state of atoms of reactant and product side involved during the reaction.**

**Is H**_{2}SO_{4} + Fe_{2}S_{3} a precipitation reaction

_{2}SO

_{4}+ Fe

_{2}S

_{3}a precipitation reaction

**H _{2}SO_{4 }+ Fe_{2}S_{3} reaction is not a precipitation reaction because no precipitates are formed during the reaction.**

**Is H**_{2}SO_{4} + Fe_{2}S_{3} a reversible or irreversible reaction

_{2}SO

_{4}+ Fe

_{2}S

_{3}a reversible or irreversible reaction

**H _{2}SO_{4 }+ Fe_{2}S_{3} is an irreversible reaction because the product formed do not react to give back reactants under the same condition.**

**Is H**_{2}SO_{4} + Fe_{2}S_{3} a displacement reaction

_{2}SO

_{4}+ Fe

_{2}S

_{3}a displacement reaction

**H _{2}SO_{4 }+ Fe_{2}S_{3} is a double displacement reaction, in which sulphate ion is displaced from H_{2}SO_{4} molecule and sulphur ion from Fe_{2}S_{3} molecule to form Ferric sulphate and hydrogen sulphide.**

**3H _{2}^{+}+ 3SO_{4}^{2-} + Fe_{2}^{3+}+ S_{3}^{2-} = Fe_{2} (SO_{4})_{3} + 3H_{2}S**

**Conclusion**

Sulphuric acid reacts with Ferric sulphide and gives Ferric sulphate and Hydrogen sulphide. It is a double displacement reaction and no oxidation or reduction occurs during the reaction.

Hello …. I am Dr. Tomleshkumar Deshmukh. I have completed my Ph.D.

from RTMNU. My area of research is Organo-Analytical.