Question

The decomposition of hydrogen iodide on a gold surface at 150 °C HI(g)½ H2(g) + ½ I2(g) is zero order in HI with a rate constant of 1.20×10-4 M s-1. If the initial concentration of HI is 0.373 M, the concentration of HI will be 0.0053 M after 2.64×103 seconds have passed. Based on these data, the rate constant for the reaction is _______M s-1.

Answer 1

Answer:

k = 1,39x10⁻⁴Ms⁻¹

Explanation:

For the reaction:

HI(g) → ½ H2(g) + ½ I2(g)

The zero order formula is:

[A] = [A]₀ - kt

Where [A] is the concentration in time, t, [A]₀ is initial concentration and k is rate constant.

When [A]₀=0,373M, At t=2,64x10³ s, [A] is 0,0053M

Replacing:

0,0053M = 0,373M - k×2,64x10³s

0,3677M = k×2,64x10³s

k = 1,39x10⁻⁴Ms⁻¹

I hope it helps!