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4.22   The rate constant for the decomposition of N2O5 at various temperatures
             is given below:

          

 

            Draw a graph between ln k and 1/T and calculate the values of A and
            E_a. Predict the rate constant at 30° and 50°C.

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From the above data,

T/C^{0} 0 20 40 60 80
T/K 273 293     313 333 353
1/T/K^{-1} (\times 10^{-3}) 3.66 3.41 3.19 3.0 2.83
10^{5}*K/S^{-} 0.0787 1.70 25.7 178 2140
ln\ K -7.147 -4.075 -1.359 -0.577 3.063


Slope of line  = \frac{y2-y2}{x2-x1} = -12.30\ K

According to Arrhenius equations,

Slope = -E_{a}/R
  E_{a}= 12.30 \times8.314 
           = 102.27 KJ mol^{-1}

Again,

When T = 30 +273 = 303 K and 1/T =0.0033K 
\ln k= -2.8

\therefore  k = 6.08\times 10^{-2}\ s^{-1}

When T = 50  + 273 = 323 K  and 1/T = 3.1 \times 10^{-3} K
\ln k = -0.5
\therefore k = 0.607 per sec

Posted by

manish

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