Chemical Equations
Chemical equations are the short representation of the chemical change in which reactants and products are written in symbols (in the case of an element) and formula (in the case of a compound). Here, we learn about various topics related to chemical equations.
Summary
Chemical equations are the short representation of the chemical change in which reactants and products are written in symbols (in the case of an element) and formula (in the case of a compound). Here, we learn about various topics related to chemical equations.
Things to Remember
- A chemical equation is the short representation of the chemical change in which reactants and products are written in symbols (in the case of an element) and formula (in the case of a compound).
- Balancing of the chemical equation means total atoms in the reactants (L.H.S) are equal to the total atoms in products (R.H.S).
- Partial equation method is very important because, in this method, we know how to decompose large molecule into simpler smaller molecules, how to carry out possible reactions, how to counsel nascent hydrogen, oxygen, and chlorine atom by multiplying with suitable numbers, and at last adding and subtracting all reactants and products to get balanced chemical equation.
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Chemical Equations
Introduction
It is the short representation of the chemical change in which reactants and products are written in symbols (in the case of an element) and formula (in the case of a compound).
For example- When hydrogen is burnt in air, we get water.
$$Hydrogen + oxygen \xrightarrow[air]{burnt in} Water$$
$$2H_2 + O_2 \xrightarrow[heat]{\Delta } 2H_2O$$
Essentials of a chemical equation
1) Chemical change must be true:
$$Mg + O_2 \xrightarrow[air]{burnt in} MgO_2 + Mg_3O_4 (wrong) $$
$$2Mg + O_2 \xrightarrow[]{} 2MgO (correct) $$
2) It should be written in molecular form (except metal and inert gas):
$$H + O \rightarrow H_2O (wrong)$$
$$H_2 + O_2 \rightarrow H_2O (correct)$$
3) It should be balanced i.e. total atoms in left side (reactants) are equal to total atoms in right hand side (products).
$$N_2 + 3H_2\rightarrow 2NH_3$$
Significance (importance) of chemical equance
When hydrogen is burnt in air, we get water.
$$2H_2 + O_2 \xrightarrow[heat]{\Delta } 2H_2O$$
A) Qualitatively: this equation represents language: When hydrogen is burnt in air, we get water.
B) Quantitatively: this equation represents:
- The number of molecules: Two molecules of hydrogen combines with one molecule of oxygen to five two molecules of water.
- The number of atoms: Four atoms of hydrogen combines with 2 atoms of oxygen to give 2 molecules of water.
- The number of volumes: {IF ALL THE REACTANTS AND PRODUCTS ARE IN GASEOUS STATE}Two volumes of hydrogen combines with one volume of oxygen to give 2 volumes of steam.
- The number of moles: Two moles of hydrogen combines with one mole of oxygen to give 2 moles of steam.
- The number of parts (by mass): Four parts by the weight of hydrogen combines with 32 parts by the weight of oxygen to give 36 parts by the weight of water.
Limitations of chemical equations
The limitations of chemical equations are as:
- State of reactants and products: Example- $$MgCO_3 + 2HCl \rightarrow MgCl_2 + H_2O + CO_2$$
- Concerntrataion of reactants and products: Example-$$ 2H_2 + O_2 \rightarrow 2H_2O$$
- The rate of reaction and time required to complete the reaction.
- The condition of the reaction: Example- $$2Mg + O_2 \rightarrow 2MgO$$
- The exothermic (heat out) and endothermic (heat in) nature of areaction: Example- $$N_2 + 3H_2\rightarrow 2NH_3$$
- The reversibility of a reaction: Example- $$N_2 + 3H_2\rightarrow 2NH_3$$
To over come these limitations following remedies were employed:
- Physical states of reactants and products are indicated by putting the letters (s), (l), (g), and (aq.) after solid, liquid, gas, and aqueous respectively. Example- $$CaCO_3 (s) + 2HCl(aq.) \rightarrow CaCl_2(s) + H_2O(l) + CO_2 (g)$$
- Condition of the reaction can be shown by writing brief information above or below the sign of arrow or equality. Example- $$2Mg + O_2 \xrightarrow[air]{burnt in} 2MgO$$
- Reversible reaction are shown by a sign of reversibility in place of single headed arrow. Example- $$N_2 + 3H_2 \rightleftharpoons 2NH_3$$
- To represent concentrated and dilute solutions, conc. and dilute or aqueous terms are used respectively below the corresponding chemicals. Example- $$MgCO_3 + 2HCl(aq.)\rightarrow MgCl_2(aq.) + H_2O + CO_2$$ $$Zn + 2H_2SO_4 (conc.) \overset{\Delta }{\rightarrow} ZnSO_4 + 2H_2O +SO_2$$
- Evolution of heat can be denoted with (+) sign at the side of products. In case of absorption (-) sign can be used. Example-$$2S0_2 + O_2 \rightarrow 2SO_3 + 42K (Exothermic reaction)$$ $$N_2 + O_2 \rightarrow 2NO - 43K (Endothermic reaction)$$
- Upward arrow is used for evolution of gas and downward arrow is used for precipitate formation. Example- $$NaCl(aq.) + AgNO_3 (aq.) \rightarrow AgCl (\downarrow) + NaNO_3(aq.) $$ $$CaCO_3 + 2HCl (dil.) \rightarrow CaCl_2 (aq.) + H_2O(l) + CO_2 (g) \uparrow$$
Balancing a Chemical Equation
Balancing of chemical equation means total atoms in the reactants (L.H.S) are equal to total atoms in products (R.H.S). For balancing a chemical equation, we discuss two methods:
⇒Hit and trail method: This method is also called trial and error method. This method is very useful for simple chemical equation.
Example : $$H_2 + Cl_2 \xrightarrow{sunlight} 2HCL$$
$$NaNO_2 + NH_4Cl \xrightarrow {\Delta} NaCl + H_2O +N_2$$
Reactant side | Product side |
No. of Na = 1 | No. of Na = 1 |
No. of N = 2 | No. of = 2 |
No. of O = 2 | No. of O =1 |
No. ofH = 4 | No. of H = 2 |
No. of Cl = 1 | No. of Cl = 1 |
In this equation, Na, N and Cl atoms are balanced. Hence, we multiply by 2 in water to get balenced reaction like:
$$NaNO_2+NH_4Cl \xrightarrow{} NaCl + 2H_2) + N_2$$
This is a balacned chemical equation.
⇒Partial equation method: This method is very important because in this method, we know how to decompose large molecule into simpler smaller molecules, how to carry out possible reactions, how to concel nascent hydrogen, oxygen and chlorine atom by multiplying with suitable numbers, and at last adding and subtracting all reactanats and products to get balanced chemical equation. This step is completed in 5 steps:
1) Step one - Large molecule is decomposed into two or more simple molecules.
Such as:
a) $$2KMnO_4 \xrightarrow {} K_2O+ MnO+ 5[O]$$
b) $$K_2Cr_2O_7 \xrightarrow {} K_2O+ Cr_2O_3 + 3[O] $$
c) $$H_2O_2 \xrightarrow {} H_2O + [O]$$
d) $$ O_3 \xrightarrow {} O_2 + [O]$$
e) $$conc. H_2SO_4 \xrightarrow {} H_2O + SO_2 + [O]$$
Nitric acid at different conditions:
I) Dilute and cold nitric acid $$ 2HNO_3 \xrightarrow {} H_2O + N_2O + 4[O]$$
II) Moderately conc. nitric acid $$ 2HNO_3 \xrightarrow {1:1} H_2O + 2NO + 3[O]$$
III) Hot and conc. nitirc acid $$ 2HNO_3 \xrightarrow{} H_2O + @NO_2 + [O]$$
2) Step two -Possible reactions are carried out :
-Acid-base reaction and - Oxidation and reduction
3) Step three -Balance all simple chemical equations in reactants and products by Hit and trail method.
4) Step four - Cancel the nascent (H,O,Cl) and unwanted compound by multiplying with asuitable number.
5) Step five -At last, add and subtract the reactants and products will give us the balanced reaction.
Reference
Adhikari, Rameshwar; Khanal, Santosh; Subba , Bimala; Adhikari, Santosh; Khatiwada, Shankar Pd. Universal Chemistry XI. First. Vol. 1st. Kathmandu: Oasis Publication, 2069.
Chaudhary, Ganga Ram; Karna, Shila Kant Lal; Sharma, Kanchan; Singh, Sanjay; Gupta, Dipak Kumar. A Textbook of Higher Secondary Chemistry XI. Ed. 2nd. Kathmandu: Vidyarthi Pustak Bhandar, 2069 (2012).
Lesson
Language of Chemistry
Subject
Chemistry
Grade
Grade 11
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