J.R. S. answered 03/07/18
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IBr(g) -> I(g) + Br(g)
IBr(g) + Br(g) -> I(g) + Br2(g)
IBr(g) + Br(g) -> I(g) + Br2(g)
I(g) + I(g) -> I2(g)
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2IBr ==> Br2 + I2
overal order = 2nd order