Question

किसी अभिक्रिया A→ उत्पाद के लिए k=2.0x10-2s-1 है। यदि A की प्रारंभिक सांद्रता 1.0mol L-1 हो तो 100s के पश्चात् इसकी सांद्रता क्या रह जाएगी?

Answer 1

Answer:

Electrolytes are substances which, when dissolved in water, break up into cations (plus-charged ions) and anions (minus-charged ions). We say they ionize. Strong electrolytes ionize completely (100%), while weak electrolytes ionize only partially (usually on the order of 1–10%). That is, the principal species in solution for strong electrolytes are ions, while the principal specie in solution for weak electrolytes is the un-ionized compound itself.

Strong electrolytes fall into three categories: strong acids, strong bases, and salts. (Salts are sometimes also called ionic compounds, but really strong bases are ionic compounds as well.) The weak electrolytes include weak acids and weak bases.

Examples of strong and weak electrolytes are given below:

Strong Electrolytes strong acids HCl, HBr, HI, HNO3, HClO3, HClO4, and H2SO4

strong bases NaOH, KOH, LiOH, Ba(OH)2, and Ca(OH)2

salts NaCl, KBr, MgCl2, and many, many more

Weak Electrolytes

weak acids HF, HC2H3O2 (acetic acid), H2CO3 (carbonic acid), H3PO4 (phosphoric acid), and many more

weak bases NH3 (ammonia), C5H5N (pyridine), and several more, all containing "N"

Being Able to Classify Electrolytes Is Critical

As chemists, we need to be able to look at a formula such as HCl or NaOH and quickly know which of these classifications it is in, because we need to be able to know what we are working with (ions or compounds) when we are working with chemicals. We need to know, for example, that a bottle labeled "NaCN" (a salt) rea

Answer 2

100s के पश्चात् [A] की  सांद्रता , $[A] = 0.135 mol L^{-1}$

Explanation:

चूकि ,  वेग स्थिरांक की इकाई $s^{-1}$ है , इसलिए दी हुई अभिक्रिया प्रथम कोटि की है |

                       $\therefore$              $k = \frac{2.303}{t} log\frac{[A]_\circ}{[A]}$

                 $\Rightarrow$     $2.0 \times 10^{-2} s^{-1} = \frac{2.303}{100 s} log\frac{ 1.0 }{[A]}$

                 $\Rightarrow$     $2.0 \times 10^{-2} s^{-1} = \frac{2.303}{100 s} (- log[A] )$

                 $\Rightarrow$     $log[A] = \frac{-2.0 \times 10^{-2} \times100 s}{2.303}$

                  $\Rightarrow$     $[A] =Antilog \left ( \frac{-2.0 \times 10^{-2} \times100 s}{2.303} \right ) = 0.135 mol L^{-1}$

इसलिए  ,  100s के पश्चात् [A] की  सांद्रता , $[A] = 0.135 mol L^{-1}$