The decomposition of formic acid on gold surface follows first order kinetics- If the rate constant at 300 K is 1-0 - 10-3 s-1 and the activation energy Ea - 11-488 kJ mol-1- the rate constant at 200 K is - 10-5 s-1-Round off to the nearest Integer-

By using Arrhenius equation, logK_2K_1 = E_a2.303R[1T_1 1T_2 ] Where, K_2 is rate constant at T_2 ( 300 K) K_1 is rate constant at T_1 (200 K) R is univers ...

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Standard XII

Chemistry

Zero Order Reaction

The decomposi...

Question

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at $300 K$ is $1.0 \times 10^{- 3} s^{- 1}$ and the activation energy $E_{a} = 11.488 {kJ mol}^{- 1}$ , the rate constant at $200 K$ is $\times 10^{- 5} s^{- 1}$ .(Round off to the nearest Integer).

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Solution

By using Arrhenius equation,
$log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] Where, K_{2} is rate constant at T_{2} (300 K) K_{1} is rate constant at T_{1} (200 K) {R is universal gas constant = 8.314 J K}^{- 1} {mol}^{- 1} Substituting the values, we get log \frac{1.0 \times 10^{- 3} s^{- 1}}{K_{1}} = \frac{11.488 \times 1000}{2.303 \times 8.314} [\frac{1}{200} - \frac{1}{300}] log \frac{10^{- 3}}{K_{1}} = 600 \times \frac{3 - 2}{600} log \frac{10^{- 3}}{K_{1}} = 1 \Rightarrow 10 = \frac{10^{- 3}}{K_{1}} \Rightarrow K_{1} = 10^{- 4} s^{- 1} S o, x \times 10^{- 5} = 10^{- 4} \Rightarrow x = 10$

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The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0×10−3 s−1 and the activation energy Ea=11.488 kJ mol−1, the rate constant at 200 K is ×10−5 s−1.(Round off to the nearest Integer).

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at $300 K$ is $1.0 \times 10^{- 3} s^{- 1}$ and the activation energy $E_{a} = 11.488 {kJ mol}^{- 1}$ , the rate constant at $200 K$ is $\times 10^{- 5} s^{- 1}$ .(Round off to the nearest Integer).