1.

The rate law for a reaction is found to be Rate = K [NO-2] [I-] [H+]2 How would the rate of reaction change when (i) Concentration of H+ is doubled (ii) Concentration of I- is halved (iii) Concentration of each of NO2, I- and H+ are tripled?

Answer»

Suppose initially the concentration are [NO-2] = a mol L-1, [I-] = b mol L-1 and [H+ ] = c mol L-1 

∴ Rate = K abc2 

(i) New [H+] = 2c 

∴ New rate = 4ab (2c)2 = 4abc2 

= 4 times 

(ii) New [I-] = \(\frac{b}{2}\)

New Rate = Ka \(\frac{b}{2}\)c2 = \(\frac{1}{2}\)Kabc2 

ie. rate of reaction is halved 

(iii) New [NO-2] = 3a, [I-] = 3b, [H+] = 3c 

New rate = K (3a) (3b) (3c)2 

= 81K abc2 = 81 times.



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