The equilibrium constant for the following reaction of hydrogen gas and bromine gas at 25°C is 5.628 X 1018?
The equilibrium constant for the following reaction of hydrogen gas
and bromine gas at 25°C is 5.628 X 1018.
H2(g) + Br2(g) <-> 2HBr(g)
Assume that equimolar amounts of H2and Br2 were present at
the beginning. Calculate the equilbrium concentration of H2if
the concentration of HBr is 0.500 M.
The answer is 2.11x10^-10 M. How do I get to this answer?
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Reaction equation:
H₂(g) + Br₂(g) ⇄ 2 HBr(g)
So the equilibrium equations are relates as:
Kc = [HBr]² / [H₂]∙[Br₂]
According to reaction of mole of hydrogen reacts with one mole of bromine, that means they undergo the the same change in concentration. When you start the reaction with an equimolar mixture, i.e. the same concentration, the concentration of H₂ and Br₂ is the same at any extent of the reaction:
[H₂] = [Br₂]
Hence,
Kc = [HBr]² / [H₂]²
=>
[H₂] = [HBr] / √Kc
= 0.500 M / √( 5.628×10¹⁸)
= 2.11×10⁻¹⁰
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