Question

From the following data; the activation energy for
the reaction (Cal/mol) H, + 12 + 2HI
Tin K)
769
1/T(in K)
1.3x10-3
1.5 x10
log.ok
2.9
1.1
667
1) 4 x 104
(3) 8 x 104
(2) 2 x 104
(4) 3 x 104​

Answer 1

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