# Chemical Equation

## A Few Examples

NaOH + HCl NaCl + H_{2}O means that (1) 1 mole of NaOH reacts with 1 mole of HCl to form 1 mole of NaCl and 1 mole of H_{2}O, (2) 40 g (that is, **molecular weight**) of NaOH react with 36.5 g of HCl to form 58.5 g of NaCl and 18 g of H_{2}O, or (3) 6.02 × 10^{23} molecules (1 mole) of NaOH react with 6.02 × 10^{23} molecules of HCl to form 6.02 × 10^{23} of NaCl and 6.02 × 10^{23} of H_{2}O. Notice that on both sides of the equation, we have one **chlorine** atom, two **hydrogen** atoms, one **oxygen** atom, and one **sodium** atom. This equation, then, is properly balanced.

For the reaction between permanganate (MnO_{4}) ion and ferrous (Fe ^{2+}) ion in an acid solution, an expression is given like this, KMnO_{4} + FeSO_{4} + H_{2}SO_{4} Fe_{2}(SO_{4})_{3} + K_{2}SO_{4} + MnSO_{4} + H_{2}O. Obviously this equation is not balanced. To remedy this, first, the equation can be rewritten as MnO + Fe^{2+} + H^{+} 4 Fe^{3+} + Mn^{2+} + H_{2}O if one recognizes that potassium (K^{+}) and sulfate (SO_{4}) ions do not enter into the reaction. Secondly, the oxidation number of manganese (Mn) is changed from +7 in MnO_{4}^{-} to +2 in Mn^{2+}, that is, Mn gains 5 electrons during the reaction. Similarly, one electron is lost from Fe^{2+} to Fe^{3+}. To make the number of electrons lost from one substance equal to the number of electrons gained by another in oxidation-reduction reactions, we need to use the least common multiple of 1 and 5, which is 5. So we have MnO 2+ + 4 + 5Fe + H 5Fe^{3+} + Mn^{2+} + H_{2}O. Thirdly, the equation has to be balanced for the number of atoms of individual elements, too. Thus a final expression is obtained as MnO_{4}^{-} + 5Fe^{2+} + 8H^{+} 5Fe^{3+} + Mn^{2+} + H_{2}O. Lastly we can add the potassium and sulfate back into the complete equation, 2KMnO_{4} + 10FeSO_{4} + 8H_{2}SO_{4} 5Fe_{2}(SO_{4})_{3} + K_{2}SO_{4} +2MnSO_{4} + 8H_{2}O. At this stage, we do not have to worry about charge balances, but atom conservation needs to be checked again and corrected.

Derivation of equations for oxidation-reduction reactions sometimes can be simplified by using a series of half reactions, whose expressions can be found in special tables of many textbooks. For example, with half-reactions of Zn Zn^{2+} + 2e and Fe^{2+} + 2e Fe, by summing them up we can obtain the equation, Zn + Fe^{2+} Zn^{2+} + Fe. Since 2e is found both on the right and left sides of the equations and does not react with anything else, it can be dropped from the combined equation.

For those reactions in which we are interested for their heats of reaction, knowing how to derive the final equations from relevant formation reactions is very useful. For example, when the formation reactions at temperature of 77°F (25°C; 298K) are given as (1) C(s) + O_{2}(g) → CO_{2}(g), Δ H_{f}° = -94,501 cal, (2) C(s) + 0.5 O_{2}(g) → CO(g), Δ H_{f}° = -26,416 cal, and (3) H_{2}(g) + 0.5 O_{2}(g) → H_{2}O, Δ H_{f}° = -57,798 cal, we can obtain the equation, CO_{2}(g) + H_{2}(g) → CO(g) + H_{2}O(g), Δ H_{f}° = 9,837 cal, by reversing (1), that is, CO_{2}(g) → C(s) + O_{2}(g), Δ H_{f}° = 94,501 cal, and adding it to (2) and (3). Therefore, the result shows an endothermic reaction with the heat of reaction of 9,837 cal at 77°F (25°C; 298K).

## Additional topics

Science EncyclopediaScience & Philosophy: *Ephemeris* to *Evolution - Historical Background*Chemical Equation - Conventions And Symbols, A Few Examples - Applications