# Find order of reaction with respect to A and B and find the rate law expression for 2A+B -> C [closed]

## closed as off-topic by Avnish Kabaj, andselisk, Tyberius, Jon Custer, MithoronDec 20 '18 at 17:33

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• 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 – YUSUF HASAN Dec 20 '18 at 7:56
• Can you please give more information, this is little to begin with – Pallavi Singh Dec 20 '18 at 8:00
• 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 – 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 : $$r=k[A]^{1.5}[B]^{0}=k[A]^{1.5}$$