Question

Consider the reaction: $H_2(g) + I_2(g) ightarrow 2HI(g)$ Rate constants have been determined at two temperatures: egin{tabular}{|c|c|c|} hline Exp.&Temp&k ($M^{-1}s^{-1}$)\ hline 1&599 K&$5.4 imes 10^{-4}$ \ 2&683 K&$2.8 imes 10^{-2}$ \ hline end{tabular} a. Calculate the activation energy for the reaction. $E_a = $\ b. Calculate the value of the rate constant at 500. K. $k = $

          Consider the reaction:
$H_2(g) + I_2(g) 
ightarrow 2HI(g)$
Rate constants have been determined at two temperatures:

egin{tabular}{|c|c|c|}
hline
Exp.&Temp&k ($M^{-1}s^{-1}$)\
hline
1&599 K&$5.4 	imes 10^{-4}$ \
2&683 K&$2.8 	imes 10^{-2}$ \
hline
end{tabular}
a. Calculate the activation energy for the reaction. $E_a = $\
b. Calculate the value of the rate constant at 500. K. $k = $

Consider the reaction:
H2(g) + I2(g)
ightarrow 2HI(g)
Rate constants have been determined at two temperatures:

egintabular|c|c|c|
hline
Exp. Temp k (M^-1s^-1)hline
1 599 K 5.4 imes 10^-4 2 683 K 2.8 imes 10^-2 hline
endtabular
a. Calculate the activation energy for the reaction. Ea =b. Calculate the value of the rate constant at 500. K. k =

Added by Jessica J.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

04/29/2022

Step 1

4 x 10^4 M^-1 s^-1 Given rate constant at 683 K (k2) = 28 x 10^-2 M^-1 s^-1 Using the Arrhenius equation: ln(k2/k1) = Ea/R * (1/T1 - 1/T2) Substitute the values: ln(28 x 10^-2 / 5.4 x 10^4) = Ea / 8.314 * (1/599 - 1/683) Solve for Ea: Ea = 160,000 J/mol Show more…

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Consider the reaction: H2 (g) + I2 (g) → 2 HI (g) Rate constants have been determined at two temperatures: Exp 1 Temp 599 K k (M-1 s-1) 5.4 x 10^-4 Exp 2 Temp 683 K k (M-1 s-1) 2.8 x 10^-2 a. Calculate the activation energy for the reaction. Ea = b. Calculate the value of the rate constant at 500. K. k =