For a decomposition of azoisopropane at 270° C it was found that at t = 0, the total pressure was 33.15 mm of Hg and after 3 minutes the total pressure was found to be 46.3 mm of Hg. Calculate the value of K for this reaction. ( CH 3)2 CHN = NCH ( CH 2)2 arrow N 2+ C 6 H 14

Question

For a decomposition of azoisopropane at 270° C it was found that at t = 0, the total pressure was 33.15 mm of Hg and after 3 minutes the total pressure was found to be 46.3 mm of Hg. Calculate the value of K for this reaction. ( CH 3)2 CHN = NCH ( CH 2)2 arrow N 2+ C 6 H 14 (A) 0.168 min- 1 (B) 0.173 min- 1 (C) 0.18 min- 1 (D) 0.154 min- 1

Tardigrade Admin · Accepted Answer

k=(2.303/t)log(po (. n - 1 .)/n po - p1) k=(2.303/3)log(33.15 (. 2 - 1 .)/2 × 33.15 - 46.3) k=0.168 min- 1

0.168 $mi n^{- 1}$

0.173 $mi n^{- 1}$

0.18 $mi n^{- 1}$

$0.154$ $mi n^{- 1}$

$k = \frac{2.303}{t} l o g \frac{p _{o} ( n - 1 )}{n p _{o} - p _{1}}$

$k = \frac{2.303}{3} l o g \frac{33.15 ( 2 - 1 )}{2 \times 33.15 - 46.3}$

$k = 0.168$ $mi n^{- 1}$

Q. For a decomposition of azoisopropane at 270∘C it was found that at t = 0, the total pressure was 33.15 mm of Hg and after 3 minutes the total pressure was found to be 46.3 mm of Hg. Calculate the value of K for this reaction. (CH3​)2​CHN=NCH(CH2​)2​→N2​+C6​H14​

0.168 min−1

0.173 min−1

0.18 min−1

0.154 min−1

k=t2.303​lognpo​−p1​po​(n−1)​ k=32.303​log2×33.15−46.333.15(2−1)​ k=0.168 min−1

0.168 $mi n^{- 1}$

0.173 $mi n^{- 1}$

0.18 $mi n^{- 1}$

$0.154$ $mi n^{- 1}$

$k = \frac{2.303}{t} l o g \frac{p _{o} ( n - 1 )}{n p _{o} - p _{1}}$

$k = \frac{2.303}{3} l o g \frac{33.15 ( 2 - 1 )}{2 \times 33.15 - 46.3}$

$k = 0.168$ $mi n^{- 1}$