he rate constant for the formation of hydrogen iodide from the elements H2(g) + I2(g) →...
he rate constant for the formation of hydrogen iodide from the elements H2(g) + I2(g) → 2HI(g) is 2.7 × 10–4 L/(mol∙s) at 600 K and 3.5 × 10–3 L/(mol∙s) at 650 K. Find the activation energy Ea. J/mol Then calculate the rate constant at 684 K. L/(mol•s)
Homework Answers
1)
Given:
T1 = 600 K
T2 = 650 K
K1 = 2.7*10^-4 L/mol.s
K2 = 3.5*10^-3 L/mol.s
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(3.5*10^-3/2.7*10^-4) = ( Ea/8.314)*(1/600 - 1/650)
2.5621 = (Ea/8.314)*(1.282*10^-4)
Ea = 166150 J/mol
Answer: 1.66*10^5 J/mol
2)
Given:
T1 = 600 K
T2 = 684 K
K1 = 2.7*10^-4 L/mol.s
Ea = 166150 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/2.7*10^-4) = (166150.0/8.314)*(1/600 - 1/684.0)
ln(K2/2.7*10^-4) = 19984*(2.047*10^-4)
K2 = 1.614*10^-2 L/mol.s
Answer: 1.6*10^-2 L/mol.s
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