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On doubling the concentration of reactant, the rate of reaction triples. Find the reaction order.

written 2.9 years ago by RakeshBhuse • 3.2k

•   modified 2.9 years ago

On doubling the concentration of reactant, the rate of reaction triples. Find the reaction order.

$$\mathrm {-r_A=k~C^n_A}$$

advanced engineering chemistry

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written 2.9 years ago by RakeshBhuse • 3.2k

Solution :

At $\mathrm{C}_{\mathrm{A1}}$, the rate is

$$-\mathrm {\mathrm r_{A1}=k C_{A1}^{n}}$$

At $\mathrm{C}_{\mathrm{A} 2}$, the rate is

$$-\mathrm{r}_{\mathrm{A2}}=\mathrm{kC}_{\mathrm{A2}}^{\mathrm{n}}$$

If

$$-r_{\mathrm{A}_{2}}=\mathrm{kC}_{\mathrm{A} 2}^{\mathrm{n}}$$

$$\mathrm {C_{A2}}=2 \mathrm {C_{A1}}$$ . we have

$$\mathrm{-r_{A2}=3\left(-r_{A1}\right)}$$

$$\mathrm {\frac{-r_{A 2}}{-r_{A 1}}=\frac{k C_{A 2}^{n}}{k C_{A 1}^{n}}}$$

$$\mathrm{\frac{-r_{A 2}}{-r_{A 1}}=\frac{C_{A 2}^{n}}{C_{A 1}^{n}}}$$

$$\frac{3\left(\mathrm{-r_{A 1}}\right)}{\mathrm{-r_{A 1}}}=\frac{\left(2\mathrm {C_{A 1}}\right)^{\mathrm n}}{\left(\mathrm {C_{A 1}}\right)^{\mathrm n}}$$

$$\mathrm {3=2^n}$$

$$\ln 3=\mathrm{n} \ln 2$$

$$1.0986=\mathrm{n}(0.693)$$

$$\mathrm n=1.6$$

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