From the following data; the activation energy forthe reaction (Cal/mo

1.	From the following data; the activation energy forthe reaction (Cal/mol) H, + 12 + 2HITin K)7691/T(in K)1.3x10-31.5 x10log.ok2.91.16671) 4 x 104(3) 8 x 104(2) 2 x 104(4) 3 x 104
Answer» COMPLETE question: From the following data, the activation energy for the REACTION is (cal/mol): H₂ + I₂ → 2HI The table given in the question is attached below. 1) 4 × 10⁴ 2) 2 × 10⁴ 3) 8 × 10⁴ 4) 3 × 10⁴ Answer: The activation energy for the reaction is 1) 4 × 10⁴ cal/mol Explanation: The activation energy is given by the FORMULA: log K = log A - Eₐ/RT On substituting the values of first row, we get, 2.9 = log A - Eₐ/(2.303R × 769) → (equation 1) On substituting the values of second row, we get, 1.1 = log A - Eₐ/(2.303R × 667) → (equation 2) On subtracting equation (2) from (1), we get, 1.8 = (Eₐ × (769 - 667))/(2 × 667 × 769 × 2.303) ∴ Eₐ = 4.17 × 10⁴ cal/mol

From the following data; the activation energy forthe reaction (Cal/mol) H, + 12 + 2HITin K)7691/T(in K)1.3x10-31.5 x10log.ok2.91.16671) 4 x 104(3) 8 x 104(2) 2 x 104(4) 3 x 104

Answer»

COMPLETE question:

From the following data, the activation energy for the REACTION is (cal/mol):

H₂ + I₂ → 2HI

The table given in the question is attached below.

1) 4 × 10⁴

2) 2 × 10⁴

3) 8 × 10⁴

4) 3 × 10⁴

Answer:

Explanation:

The activation energy is given by the FORMULA:

log K = log A - Eₐ/RT

On substituting the values of first row, we get,

2.9 = log A - Eₐ/(2.303R × 769) → (equation 1)

On substituting the values of second row, we get,

1.1 = log A - Eₐ/(2.303R × 667) → (equation 2)

On subtracting equation (2) from (1), we get,

1.8 = (Eₐ × (769 - 667))/(2 × 667 × 769 × 2.303)

∴ Eₐ = 4.17 × 10⁴ cal/mol