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Chemistry Equilibrium Question ! Help please.?
The equation is H2 + I2 <-> 2HI
There are four pie graphs from I to IV, with pie graph I having the least shaded area, then II, then III, and IV has the most shaded area.
These shaded area represent the volume of I2 in the system at a certain unknown temperature
Which one of the pie graphs have the largest Keq?
Thanks!
A large Keq value denotes that the forward reaction favours the products moreso than the reverse reaction does.
You would have to determine if this reaction is endothermic or exothermic, so you can place the the thermal energy term as a reactant or product. Using Le Chatelier’s principles, you can then determine whether or not an increase in heat will yield more product, or if a decrease will.
Hopefully you have an understanding of thermochemical equations and enthalpy changes.
If not, research it.
To start you out, you are dealing with a synthesis reaction. I believe most synthesis reactions are exothermic, meaning that an energy value is placed as a product for the forward reaction (although I could be wrong). Following Le Chatelier’s principles, an increae in a reactant will increase the yield of a product; therefore, if you add more heat, more I2 should come out; equilibrium shifts towards the reverse reaction’s products. That means pie graph IV would correspond to the highest temperature if this is the case.
Hope this helps.
Matt | Feb 08, 2010
The largest Keq will be the one with the SMALLEST shaded area, or smallest I2 concentration since:
Keq = { [HI]^2 } / { [H2] x [I2] }
and the the smallest the denominator, the larger the fraction!
PS: You’ll have to assume that the conc of the other reagent – product did not change much!
ATorres | Feb 08, 2010
