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  • The rate constant for the decomposition of N_2 O_5 at various temperatures is given below:

The rate constant for the decomposition of N_2O_5 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.

Answers (1)

best_answer

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


The slope of line   =\frac{y 2-y 2}{x 2-x 1}=-12.30 \mathrm{~K}

According to Arrhenius's equations,

\begin{aligned} \text { Slope } & =-E_a / R \\ E_a & =12.30 \times 8.314 \\ & =102.27 \mathrm{~K} \mathrm{Jmol}^{-1}\end{aligned}

Again,

\begin{aligned} & \ln k=\ln A-\frac{E_a}{\mathrm{R} T} \\ & \ln A=\ln k+\frac{E_a}{\mathrm{R} T}\end{aligned}

WhenT=273 \mathrm{~K},
\ln k=-7.147

$
\text { Then, } \begin{aligned} \ln A & =-7.147+\frac{102.27 \times 10^3}{8.314 \times 273} \\ & =37.911 \end{aligned}
 

Therefore, A=2.91 \times 10^6


When T = 30 +273 = 303 K and 1/T =0.0033K 

\begin{aligned} & \ln k=-2.8 \\ & \therefore \mathrm{k}=6.08 \times 10^{-2} \mathrm{~s}^{-1}\end{aligned}

 

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

Posted by

manish

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