Question

For a first order reaction, if the initial concentration of the reactant is 0.8 M and rate constant is 6.93 × 10-³ s, then the rate of the reaction after 100 s will be​

Answer 1

Answer:

1.9 x 10-4 M s 1

Explanation:

Answer 2

Answer:

Rate of reaction after 100s would be

$\frac{dA}{dt} =kC_{2}=6.93\times10^{-3}(0.4)=2.772\times10^{-3}molL^{-1}s^{-1}$

Explanation:

Given that,

For a first order reaction,

The initial concentration of the reactant is $C_{1}=0.8M$ and

The rate constant of the reaction is $k=6.93\times10^{-3}s^{-1}$

We know that,

For a first order reaction the half life is given by the relation

$t_{1/2}=\frac{0.693}{k}$

Substituting the known value in above relation we get

$t_{1/2}=\frac{0.693}{6.93\times10^{-3}}=100s$

Half life of reactant is 100s.Therefore,

Concentration of reactant after 100seconds is $C_{2}=\frac{0.8}{2} =0.4M$

Therefore,

Rate of reaction after 100s would be

$\frac{dA}{dt} =kC_{2}=6.93\times10^{-3}(0.4)=2.772\times10^{-3}molL^{-1}s^{-1}$

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