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Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

The rate constant (k) for a reaction was measured as a function of temperature. ?

A plot of lnk versus 1/T(in K) is linear and has a slope of −1.29×104 K . Calculate the activation energy for the reaction.

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  • Dr W
    Lv 7
    2 months ago

    anytime you read.. "rate constant"... "activation energy"... "temperature" in the same problem statement, you should immediately think "arrhenius equation"

    .. k = Ao exp(-Ea / RT)

    if we consider 2 conditions.. where k and T vary, we can write

    .. k1 = Ao exp(-Ea / RT1)

    .. k2 = Ao exp(-Ea / RT2)

    dividing the second by the first

    .. (k2 / k1) = (Ao / Ao) * exp(-Ea/RT2) / exp(-Ea / RT1)

    rearranging a bit and simplifying... (note that a^b / a^c = a^(b-c))

    .. k2 / k1 = exp( (-Ea/R) * (1/T2 - 1/T1) )

    taking ln of both sides to eliminate exp

    .. ln(k2 / k1) = (-Ea/R) * (1/T2 - 1/T1)

    rearranging.. recall that ln(a/b) = ln(a) - ln(b)

    .. ln(k2) - ln(k1) = (-Ea/R) * (1/T2 - 1/T1)

    rearranging

    .. ln(k2) = (-Ea/R) * (1/T2) + [(Ea/R)*(1/T1) + ln(k1)]

    and if we say we're going to START with a FIXED value of T and it's corresponding k.. (let's call those To and ko) and then vary T and measure k, we can rewrite that equation as

    .. ln(k) = (-Ea/R) * (1/T) + [(Ea/R)*(1/To) + ln(ko)]

    and if you notice.. all that [(Ea/R)*(1/To) + ln(ko)] stuff on the right is a constant so that

    .. ln(k) = (-Ea/R) * (1/T) + C

    **********

    since Ea and R are also constants, 

    .. ln(k) = (-Ea/R) * (1/T) + C

    is LINEAR....

    .. y = mx + b 

    right?

    meaning a plot of  

    .. ln(k) on the y-axis

    .. (1/T) on the x-axis

    will have slope = -Ea/R

    *********

    *********

    got all that under control?  back to this problem

    .. -Ea/R = -1.29x10^4 K

    .. Ea = R * 1.29x10^4 K

    .. Ea = 8.314 J/molK * 1.29x10^4 K

    .. Ea = 107 kJ/mol

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  • 2 months ago

    This is a plot of a linear form of the Arrhenius equation. In this form of the equation, the slope of this line is equal to - Ea/R. So,

    -1.28X10^4 K = -Ea / 8.314 J/molK

    Ea = 1.06X10^5 J/mol = 106 kJ/mol

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