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For a chemical raction AtoB the rate of the reaction is 2xx10^(-3) mol dm^(-3)s^(-1), when the initial concentration is 0.05 mol dm^(-3). The rate of the same reaction is 1.6xx10^(-2) mol dm^(-3)s^(-1) when the initial concentration is 0.1 mol dm^(-3). The order of the reaction is |
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Answer» 0 rate `=k[A]^(n)` where `k=` rate CONSTANT, n=order of reaction [A]= concenration of reactant given`"rate"_(1)=2xx10^(-3)"mol dm^(-3)s^(-1)` `[A_(0)'=0.05"mol"dm^(-3)` `"rate"_(II)=1.6xx10^(-2)"mol"dm^(-3)s^(-1)` `[A_(0)]=0.1"mol"dm^(-3)` `:.2xx10^(-3)=k[0.1]^(n)`...............i `1.6xx10^(-2)=k[0.1]^(n)`.............ii Divide i by ii `IMPLIES(2xx10^(-3))/(1.6xx10^(-2))=([0.05]^(n))/([0.1]^(n))=([0.05]^(n))/(2^(n)[0.05]^(n))` `implies1/(2^(n))=1/8` `impliesn=3, :.` Order of reaction =3 |
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