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The rate cosntant of first order reaction is `6.8 xx 10^(-4) s^(-1)` If the initial concentration of the reactant is 0.04 M , What is its molarity after 20 minutes ? How long will it take for 25% of the reactant to react ? |
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Answer» Correct Answer - (i) Molarity of reactant after 20 min =0.0177 M (ii) Time for 25% of the reaction = 7.05 min Given : Rate constant `=k = 6.8 xx 10^(-4) s^(-1)` `[A]_(0) =0.04 M` (i) After t= 20 min = 20 x 60 s = 1200 s `[A]_(t) = ?` `k= (2.303)/(t) log_(10) .([A]_(0))/([A]_(t))` `:. log_(10).[[A]_(0)]/[[A]_(t)] = (kxxt)/(2.303) - (6.8 xx 10^(-4) xx 1200)/(2.303 ) = 0.3543` `:. [[A]_(0)]/[[A]_(t)]=AL 0.3543 =2.26` `:. [A]_(t) =[[A]_(0))/(2.26) =(0.04)/(2.26) =0.0177 M` (ii) For 25% reactant to react t= ? Reactant reacted =25% of 0.04 M `=(25)/(100) xx 0.04` `=0.01 M` `:.` Reactant left `=[A]_(t) = 0.04 - 0.01 =0.03 M` `t= (2.303)/(k) log_(10) .([[A]_(0)])/([[A]_(t)])` `=(2.303)/(6.8 xx 10^(-4)) log_(10) .(0.04)/(0.03)` `=(2.303)/(6.8xx 10^(-4))` `=423 s` `=(423)/(60) min = 7.05 min` |
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