At 298 K, the rate constant for a reaction is 0.0346 s-1. What is the rate constant at 350K if the Ea = 50.2kJ/mol

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At 298 K, the rate constant for a reaction is 0.0346 s-1. What is the rate constant at 350K if the Ea = 50.2kJ/mol

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Thu Nguyệt 4 years 2021-08-11T06:13:42+00:00 1 Answers 18 views 0

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  1. Xavia
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    2021-08-11T06:15:02+00:00
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    Answer:

    0.702 /s

    Explanation:

    Rate constant at [298 \mathrm{~K}, \mathrm{~K}_{1}=3.46 \times 10^{-2} \mathrm{~s}^{-1}

    Rate constant at 350 \mathrm{~K}, \mathrm{~K}_{2}=?

    T_{1}=298 \mathrm{~K}

    T_{2}=350 \mathrm{~K}

    Activation energy, \mathrm{Ea}=50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}

    Use the following equation to calculate K_{2}$ at $350 \mathrm{~K}

    Use the following equation to calculate K_{2}$ at $350 \mathrm{~K}

    \ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]

    Therefore,

     \ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right]

    \ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right]

    \frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=\mathrm{e}^{3.01}

    \frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=20.3

    K_{2} &=20.3 \times 3.46 \times 10^{-2} \mathrm{~s}^{-1}

    &=0.702 \mathrm{~s}^{-1}

    hence, the rate constant at 350 \mathrm{~K} is 0.702\mathrm{~s}^{-1}

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