Which of the following would an increase in pressure cause a shift in equilibrium that favors reactants?
Select one:
a. CO2(g)+H2(g) ⇌CO(g)+H2O(g)CO2(g)+H2(g) ⇌CO(g)+H2O(g)
b. CO(g) +12O2(g) ⇌CO2(g)CO(g) +12O2(g) ⇌CO2(g)
c. 2Hg(l)+O2(g) ⇌2HgO(s)2Hg(l)+O2(g) ⇌2HgO(s)
d. 2H2(g)+O2(g) 2H2O(l)2H2(g)+O2(g) 2H2O(l)
e. CaCO3(s) ⇌CaO(s)+CO2(g)
Answers : e.
Because in this reaction left side (reaction side) the number of gas molecular is zero,while right side of the reaction (product side)one molecular is gas.Therefore according to Le chatelier's principal ......Pressure increase when less practicals = less colliosion = less pressure.and Pressure decrese when more particals = more collision = rised pressure.
And this situation has not in other reactions which is given in option a,b,c and d.However,in this reaction the number of mole is equal in both side of the reaction.So, there pressure is zero.
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