Question 15.6: Consider the reaction at chemical equilibrium: 2 KClO3(s) ⇌ ......
Consider the reaction at chemical equilibrium:
2\operatorname{KClO_{3}}(s)\rightleftharpoons2\operatorname{KCl}(s)+3\operatorname{O}_{2}(g)What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture?
Step-by-Step
The chemical equation has 3 mol of gas on the right and 0 mol of gas on the left. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to the left (toward the side with fewer moles of gas particles). Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the right (toward the side with more moles of gas particles).
Related Answered Questions
Question: 15.8
Verified Answer:
To write the expression for K_{\mathrm{sp}}...
Question: 15.11
Verified Answer:
K_{\mathrm{eq}}={\frac{[I]^{2}}{[I_{2}]}}[/...
Question: 15.10
Verified Answer:
K_{\mathrm{eq}}={\frac{[\mathrm{NOBr}]^{2}}...
Question: 15.9
Verified Answer:
Begin by writing the reaction by which solid [late...
Question: 15.12
Verified Answer:
K_{\mathrm{eq}}={\frac{\mathrm{[NH_{3}]}^{2...
Question: 15.13
Verified Answer:
(a) Shift left
(b) Shift right
(c) Shift right (mo...
Question: 15.14
Verified Answer:
\mathrm{P b C l}_{2}(s)\rightleftarrows\mat...
Question: 15.15
Verified Answer:
\operatorname{AgI}(s)\ {\rightleftarrows\op...
Question: 15.7
Verified Answer:
Because the reaction is endothermic, we can think ...
Question: 15.5
Verified Answer:
Adding more CO_2 increases the conc...