The rate constant for the decomposition of N2O5 at various temperatures is given below- T-C 0 20 40 60 80 105- k-s-1 0-0787 1-70 25-7 178 2140Draw a graph between ln k and 1-T and calculate the values of A and Ea- Predict the rate constant at 30- and 50- C-

ln k = lnA E_aRT y = mx + c From the graph, m = 191.4 c = 4.922 Hence, ln k = 4.922 + 191.4T At 30 ^oC or 303 K, ...

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Standard XII

Chemistry

Rate Constant

The rate cons...

Question

The rate constant for the decomposition of $N_{2} O_{5}$ at various temperatures is given below:
$T /^{\circ} C$ $0$ $20$ $40$ $60$ $80$
$10^{5} \times k / s^{- 1}$ $0.0787$ $1.70$ $25.7$ $178$ $2140$
Draw a graph between ln $k$ and $1 / T$ and calculate the values of A and $E_{a}$ . Predict the rate constant at $30^{\circ}$ and $50^{\circ} C$ .

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Solution

$l n k = l n A - \frac{E_{a}}{R T}$
$y = m x + c$
From the graph,
$m = - 191.4$
$c = - 4.922$
Hence, $l n k = - 4.922 + \frac{- 191.4}{T}$
At $30$ $^{o} C$ or $303$ K,
$l n k = - 4.922 + \frac{- 191.4}{303}$
$l n k = - 5.515$
$k = 0.003873 / s$

At $50$ $^{o} C$ or $323$ K,
$l n k = - 4.922 + \frac{- 191.4}{323}$
$l n k = - 5.55$
$k = 0.004028 / s$

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$T /^{\circ} C$	$0$	$20$	$40$	$60$	$80$
$10^{5} \times k / s^{- 1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$

The rate constant for the decomposition of N2O5 at various temperatures is given below:T/∘C020406080105×k/s−10.07871.7025.71782140Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30∘ and 50∘C.

lnk=lnA−EaRTy=mx+cFrom the graph,m=−191.4c=−4.922Hence, lnk=−4.922+−191.4TAt 30 oC or 303 K,lnk=−4.922+−191.4303lnk=−5.515k=0.003873/sAt 50 oC or 323 K,lnk=−4.922+−191.4323lnk=−5.55k=0.004028/s