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On heating orisine (AsH3)) decomposes as...

On heating orisine `(AsH_3))` decomposes as :
`2AsH_(3)(g)rarr2As(s)+3H_(2)(g)`
The total pressure measured at constant temperature and constant volume varies with time as follows:
`{:(t("min"),0,5,7.5,10),(p//mmHg,760,836,866.4,896.8):}`
Calculate the rate constant assuming the reaction to follow the first order rate law.

Text Solution

Verified by Experts

For first order, the rate equation is
Initially after time t
`k'=(2.303)/(t)log.([A]_(0))/([A])=(2.303)/(t)log.(P_(0))/(p)`
given `p_(0)=760mm Hg.`
The decomposition reaction is :
`underset(P0-2x)underset(P0)(2AsH_(3))(g)rarrAsunderset(0)((s))+3H_(2)underset(3x)underset(0)((g))`
Total pressure, `p_(t)=p_(0)-2x+3x=p_(0)+x`
`x=p_(t)-p_(0)`
`P_(AsH_(3))=(p_(0)-2x)=p_(0)-2p_(t)+2p_(0)=3p_(0)-2p_(t)`
After 5 minutes , `P_(AshH_(3))=(3xx760)-(2xx836)`
=608 mm Hg
`k=(2.3030)/(5)log_(10).(760)/(608)=0.446 "min"^(-1)`
After 7.5 minutes, `p_(AsH_(3))=(3xx760)-(2xx896.8)`
`547.2 mm Hg`
`k=(2.303)/(10)log_(10).(760)/(547.2)=0.0438 "min"^(-1)`
After 10 minutes `P_(AsH_(3))=(3xx760)-(2xx896.8)`
`=486.4 mm Hg`
`k=(2.303)/(10)log_(10).(760)/(486.4)=0.00446 "min"^(-1)`
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