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Please help me in this chemistry question


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  • $\begingroup$ Try to look at two entries where either concentration of A or B is constant, and observe how the the rate and the concentrarion of the other reactant varies $\endgroup$ – YUSUF HASAN Dec 20 '18 at 7:56
  • $\begingroup$ Can you please give more information, this is little to begin with $\endgroup$ – Pallavi Singh Dec 20 '18 at 8:00
  • $\begingroup$ For example, in experiment 1 and 2, conc of A is constant. Also, you can see that as concentration of B changes, rate of reaction remains unaffected by that change. So this means that the rate would be independent of concn of B. Now similarly,try for A $\endgroup$ – YUSUF HASAN Dec 20 '18 at 8:03

1- Write the rate law expression : $r=k[A]^x[B]^y$

2-Take experiment one and two : $$5.07\times{10^{-5}}=k[0.2]^x[0.3]^y\tag{1}$$ $$5.07\times{10^{-5}}=k[0.2]^x[0.1]^y\tag{2}$$ 3- Divide equation one by equation two: $$\frac{5.07\times{10^{-5}}}{5.07\times{10^{-5}}}=\frac{k[0.2]^x[0.3]^y}{k[0.2]^x[0.1]^y}\tag{3a}$$ $$1=3^{y}\tag{3b}$$ $$3^{0}=3^{y}\tag{3c}$$ ,so:y=0(the order of reaction with respect to B), now : the rate law expression : $r=k[A]^x[B]^{0}=k[A]^x$

4-Take experiment two and three: $$5.07\times{10^{-5}}=k[0.2]^x\tag{4}$$ $$1.43\times{10^{-4}}=k[0.4]^x\tag{5}$$ 5-Divide equation five by equation four: $$\frac{1.43\times{10^{-4}}}{5.07\times{10^{-5}}}=\frac{k[0.4]^x}{k[0.2]^x}\tag{6a}$$ $$2.8=2^{x}\tag{6b}$$ $$log2.8= x log 2\tag{6c}$$,so:$x\approx{1.5}$(the order of reaction with respect to A), now : the rate law expression : $r=k[A]^{1.5}[B]^{0}=k[A]^{1.5}$


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