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प्रश्न
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
| t (sec) | P(mm of Hg) |
| 0 | 35.0 |
| 360 | 54.0 |
| 720 | 63.0 |
Calculate the rate constant.
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उत्तर
Azoisopropane decomposes according to the following equation:
\[\ce{(CH3)2CHN = NCH(CH3)2_{(g)} -> N2_{(g)} + C6H14_{(g)}}\]
This action is of first order reaction.
Initial pressure P0 = 35.0 mm Hg
Decrease in pressure of azoisopropane after time t = P
Increase in N2 pressure = `P_(N_2)`
Increase in the pressure of hexane = `P_(C_6H_14)`
Total pressure of the mixture Pt = `P_A + P_(N_2) + P_(C_6H_14)`
Pt = (P0 – P) + P + P = P0 + P
P = Pt – P0
PA = P0 – (Pt – P0) = 2P0 – Pt
But PA ∝ a − x and P0 ∝ Pt
k = `2.303/t log P_0/(2P_0 - P_t)`
When t = 360 s,
k = `2.303/360 log 35/(2 xx 35 - 54)`
= `2.303/360 log 35/16`
= `2.303/360 log 2.1875`
= `2.303/360 xx 0.339`
k = 2.17 × 10–3 s–1
When, t = 720 s,
k = `2.303/720 log 35/(2 xx 35 - 63)`
= `2.303/720 log 35/7`
= `2.303/720 (log 35 - log 7)`
= `2.303/720 (1.544 - 0.845)`
= `2.303/720 xx 0.699`
k = 2.23 × 10–3 s–1
Rate constant = `((2.17 + 2.23))/2 xx 10^-3 s^-1`
= 2.20 × 10–3 s–1
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