In the reaction, CH3COCH3(g) → C2H4(g)+H2(g)+CO(g) he initial pressure is found to be 0.40 atm and after 10 min, it was 0.50 atm. The rate constant for first order reaction is [log4 = 0.6021, log 3.5 = 0.5441]

Question

In the reaction, CH3COCH3(g) → C2H4(g)+H2(g)+CO(g) he initial pressure is found to be 0.40 atm and after 10 min, it was 0.50 atm. The rate constant for first order reaction is [log4 = 0.6021, log 3.5 = 0.5441] (A) 0.0133 min-1 (B) 0.4 s-1 (C) 10 s-1 (D) 0.6 min-1

Tardigrade Admin · Accepted Answer

CH3COOCH3(l) → C2H4(g)+H2(g)+CO(g) beginmatrix0.4&0&0&0 0.4-x&x&x&x endmatrix 0.4-x+x+x+x=0.50 ⇒ 2x=0.10 ⇒ x=0.05 k=(2.303/10) log (0.4/0.4-0.05)=(2.303/10) log (0.40/0.35) =(2.303/10) log (40/35)=(2.303/10)[1.6021-1.5441] =(2.303/10)×0.0580=(0.1335/10)=0.01335 min-1 =0.01335×60=0.8 s-1

Q. In the reaction,
$C H_{3} COC H_{3 (g)} \to C_{2} H_{4 (g)} + H_{2 (g)} + C O_{(g)}$
he initial pressure is found to be 0.40 atm and after 10 min, it was 0.50 atm. The rate constant for first order reaction is [ $l o g^{4}$ = 0.6021, log 3.5 = 0.5441]

$0.0133 mi n^{- 1}$

$0.4 s^{- 1}$

$10 s^{- 1}$

$0.6 mi n^{- 1}$

Q. In the reaction, CH3​COCH3(g)​→C2​H4(g)​+H2(g)​+CO(g)​ he initial pressure is found to be 0.40 atm and after 10 min, it was 0.50 atm. The rate constant for first order reaction is [log4 = 0.6021, log 3.5 = 0.5441]

0.0133min−1

0.4s−1

10s−1

0.6min−1

CH3​COOCH3(l)​→C2​H4(g)​+H2(g)​+CO(g)​ 0.40.4−x​0x​0x​0x​ 0.4−x+x+x+x=0.50 ⇒2x=0.10⇒x=0.05 k=102.303​log0.4−0.050.4​=102.303​log0.350.40​ =102.303​log3540​=102.303​[1.6021−1.5441] =102.303​×0.0580=100.1335​=0.01335min−1 =0.01335×60=0.8s−1

Q. In the reaction,
$C H_{3} COC H_{3 (g)} \to C_{2} H_{4 (g)} + H_{2 (g)} + C O_{(g)}$
he initial pressure is found to be 0.40 atm and after 10 min, it was 0.50 atm. The rate constant for first order reaction is [ $l o g^{4}$ = 0.6021, log 3.5 = 0.5441]

$0.0133 mi n^{- 1}$

$0.4 s^{- 1}$

$10 s^{- 1}$

$0.6 mi n^{- 1}$