2HBr(g) ↔ H2(g) + Br2(g) Initially a container contains 0.62 M HBr and no product.?
What is the equilibrium concentration of HBr if the equilibrium concentration of H2 is 0.13 M?
if you could please explain. Thanks :]
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by the equation:
2HBr(g) ↔ H2(g) + Br2(g)
every mole of H2 was produced from 2 moles of HBr
so
a concentration of H2 is 0.13 M was done so by 0.26 M HBr reacting ---> H2 & Br2
What is the equilibrium concentration of HBr
0.62 M at the begginning - 0.26 M lost = 0.36 M HBr remains