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 s-1.

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!

you need to have a picture of the atom that way we can answer it

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