Subject: Chemistry
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.
$$Hydrogen + oxygen \xrightarrow[air]{burnt in} Water$$
$$2H_2 + O_2 \xrightarrow[heat]{\Delta } 2H_2O$$
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).
$$N_2 + 3H_2\rightarrow 2NH_3$$
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 limitations of chemical equations are as:
To over come these limitations following remedies were employed:
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).
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