If the temperature coefficient $α$ $alpha$ for a certain reaction is $2.5$ $2.5$ , then if the reaction was run at $283^{\circ} C$ $283^@ "C"$ and $293^{\circ} C$ $293^@ "C"$ , what is the activation energy in $kJ/mol$ $"kJ/mol"$ ?

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If the temperature coefficient $α$ $alpha$ for a certain reaction is $2.5$ $2.5$ , then if the reaction was run at $283^{\circ} C$ $283^@ "C"$ and $293^{\circ} C$ $293^@ "C"$ , what is the activation energy in $kJ/mol$ $"kJ/mol"$ ?

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Answer 1 · 2016-12-17T05:08:06

I got about:

$E_{a} = 6.5 \times 10^{3}$ $E_a = 6.5xx10^3$ $kJ/mol$ $"kJ/mol"$

From Wikipedia, the definition of the temperature coefficient $α$ $alpha$ , in regards to the rate constant $k$ $k$ and temperature $T$ $T$ , is:

$\frac{d k}{k} = α d T$ $(dk)/(k) = alphadT$

If we integrate this from state 1 to state 2 on the left side, and $T_{1}$ $T_1$ to $T_{2}$ $T_2$ on the right side:

$\int_{(1)}^{(2)} \frac{1}{k} d k = \int_{T_{1}}^{T_{2}} α d T$ $int_((1))^((2)) 1/k dk = int_(T_1)^(T_2) alphadT$

When we assume that the temperature coefficient stays constant across this small temperature range, we can pull $α$ $alpha$ out of the integral and obtain (noting that the integral of $\frac{1}{x}$ $1/x$ is $ln x$ $lnx$ ):

$ln (k_{2}) - ln (k_{1})$ $ln(k_2) - ln(k_1)$

$= ln (\frac{k_{2}}{k_{1}}) = α (T_{2} - T_{1})$ $= color(green)(ln(k_2/k_1)) = alpha(T_2 - T_1)$

$= (2.5) (293^{\circ} C - 283^{\circ} C)$ $= (2.5)(293^@ "C" - 283^@ "C")$

$= 25$ $= color(green)(25)$

where we used the fact that intervals in the celsius temperature scale are equal to intervals on the Kelvin temperature scale.

Thus, we have indirectly calculated $ln (\frac{k_{2}}{k_{1}})$ $ln(k_2/k_1)$ . Recall that this shows up in the form of the Arrhenius equation that demonstrates the temperature dependence of the rate constant:

$ln (\frac{k_{2}}{k_{1}}) = - \frac{E_{a}}{R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$ $ln(k_2/k_1) = -E_a/R[1/T_2 - 1/T_1]$

Therefore, we can now algebraically solve for and calculate the activation energy $E_{a}$ $E_a$ , in $kJ/mol$ $"kJ/mol"$ , as:

$E_{a} = - \frac{R ln (\frac{k_{2}}{k_{1}})}{\frac{1}{T_{2}} - \frac{1}{T_{1}}}$ $color(blue)(E_a) = -(Rln(k_2/k_1))/(1/T_2 - 1/T_1)$

$= - \frac{R ln (\frac{k_{2}}{k_{1}})}{\frac{T_{1} - T_{2}}{T_{1} T_{2}}}$ $= -(Rln(k_2/k_1))/((T_1 - T_2)/(T_1T_2))$

$= - R ln (\frac{k_{2}}{k_{1}}) [\frac{T_{1} T_{2}}{T_{1} - T_{2}}]$ $= -Rln(k_2/k_1)[(T_1T_2)/(T_1 - T_2)]$

$= - (0.008314472 kJ/mol \cdot K) (25) [\frac{(556.15 K) (566.15 K)}{556.15 - 5 66.15 K}]$ $= -("0.008314472 kJ/mol"cdot"K")(25)[(("556.15 K")("566.15 K"))/(556.15 - 5"66.15 K")]$

$= {6.5}_{45} \times 10^{3}$ $= color(blue)(6.5_(45)xx10^3)$ $kJ/mol$ $color(blue)("kJ/mol")$

where the subscripts indicate digits past the last significant figure.

This means that the reactants have about a $6500-kJ$ $"6500-kJ"$ energy barrier in order for the reaction to proceed successfully.

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If the temperature coefficient $α$ $alpha$ for a certain reaction is $2.5$ $2.5$ , then if the reaction was run at $283^{\circ} C$ $283^@ "C"$ and $293^{\circ} C$ $293^@ "C"$ , what is the activation energy in $kJ/mol$ $"kJ/mol"$ ?

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If the temperature coefficient $α$ $alpha$ for a certain reaction is $2.5$ $2.5$ , then if the reaction was run at $283^{\circ} C$ $283^@ "C"$ and $293^{\circ} C$ $293^@ "C"$ , what is the activation energy in $kJ/mol$ $"kJ/mol"$ ?