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(C) The rate constant $k$ is given by the Arrhenius equation: $k = A \cdot e^{-\frac{E_a}{RT}}$.For the uncatalysed reaction,$E_{a1} = 83.314 \, kJ \,

Vedclass · Accepted Answer

$7.38$

Vedclass · Answer

(C) The rate constant $k$ is given by the Arrhenius equation: $k = A \cdot e^{-\frac{E_a}{RT}}$.For the uncatalysed reaction,$E_{a1} = 83.314 \, kJ \, mol^{-1} = 83314 \, J \, mol^{-1}$.For the catalysed reaction,$E_{a2} = 75 \, kJ \, mol^{-1} = 75000 \, J \, mol^{-1}$.The ratio of rate constants is $\frac{k_2}{k_1} = \frac{A \cdot e^{-\frac{E_{a2}}{RT}}}{A \cdot e^{-\frac{E_{a1}}{RT}}} = e^{\frac{E_{a1} - E_{a2}}{RT}}$.Given $R = 8.314 \, J \, K^{-1} \, mol^{-1}$ and $T = 500 \, K$,the difference in activation energy is $\Delta E_a = 83314 - 75000 = 8314 \, J \, mol^{-1}$.Thus,$\frac{k_2}{k_1} = e^{\frac{8314}{8.314 	imes 500}} = e^{\frac{8314}{4157}} = e^2$.Since $e \approx 2.718$,$e^2 \approx 7.389 \approx 7.38$.

$A$ catalyst lowers the activation energy for a certain reaction from $83.314 \, kJ \, mol^{-1}$ to $75 \, kJ \, mol^{-1}$ at $500 \, K$. What will be the rate of reaction as compared to the uncatalysed reaction? Assume other things being equal.

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The rate constant is given by the equation $k = p Z e^{-E_a/RT}$. Which factor should register a decrease for the reaction to proceed more rapidly?

The rate constant for the reaction,$COCl_{2(g)} \longrightarrow CO_{(g)} + Cl_{2(g)}$ is given by $\ln[k / (min^{-1})] = -11067 / T(K) + 31.33$. The temperature at which the rate of this reaction will be doubled from that at $25^{\circ} C$ is $..... \, ^{\circ} C$.

Rate constant varies with temperature by the equation $log_{10} K = 5 - 2000 / T$. We can conclude that $(R = 8.314 \ J \ mol^{-1} K^{-1})$

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