In reaction: A(l) +B(s) +C(g) <--> D(g) +E(g) Where would the reaction shift if i) B(s) is added ii) A(l) is added?
Since there is a solid and a pure liquid, will they have no effect and therefore no shift since the concentrations of both are constant?
Help with this chemistry reaction shift question?
2017-10-09 2:21 pm
回答 (1)
2017-10-09 3:41 pm
A(l) + B(s) + C(g) ⇌ D(g) +E(g)
(i)
When B(s) is added, there will NOT be any shift in equilibrium position. This is because it is a heterogeneous equilibrium and B exists as a solid phase.
(ii)
When A(l) is added, there will also NOT be any shift in equilibrium position. This is because it is a heterogeneous equilibrium and A exists as a liquid phase.
(i)
When B(s) is added, there will NOT be any shift in equilibrium position. This is because it is a heterogeneous equilibrium and B exists as a solid phase.
(ii)
When A(l) is added, there will also NOT be any shift in equilibrium position. This is because it is a heterogeneous equilibrium and A exists as a liquid phase.
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