Answer:
The reaction is shifted to the left.
Explanation:
- Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
- When there is a decrease in pressure, the equilibrium will shift towards the side with more no. of moles of gas molecules of the reaction.
- The reactants side (left) has 4.0 moles of gases and the products side (right) has 2.0 moles of gases.
- So, decreasing the pressure will shift the reaction to the side with more moles of gas (left side).
- <em>so, the reaction is shifted to the left.</em>
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If we abbreviate the formula for nicotine as Nic, then the equations for two different equilibria of Nic in water are
Nic + H2O ---> NicH+ + OH-
NicH+ + H2O ---> NicH2 2+ + OH-
We can write the Kb1 expression for the first equation as
Kb1 = 1.0×10^-6 = [NicH+][OH-] / [Nic]
1.0×10^-6 = x^2 / 1.85×10^-3 - x
Approximating that x is negligible compared to 1.85×10^-3 simplifies the equation to
1.0×10^-6 = x^2 / 1.85×10^-3
x = 0.0000430
x = [OH-] = 4.30×10^-5 M
From the Kb2 expression
Kb2 = 1.3×10-11 = [NicH2 2+][OH-] / [NicH+]
1.1×10^-10 = x^2 / 4.30×10^-5 - x
Approximating that x is negligible compared to 4.30×10^-5 simplifies the equation to
1.1×10^-10 = x^2 / 4.30×10^-5
x = [OH-] = 6.88×10^-8
The concentration [OH-] can be computed as
[OH-] = 4.30×10^-5 M + 6.88×10^-8 M = 4.30×10^-5 M
This shows that the second equilibrium has a negligible effect on the pH.
We can now calculate for pH:
pOH = -log [OH-] = -log (4.30×10^-5 M) = 4.37
pH = 14 - pOH = 14 - 4.37 = 9.63
Answer:
The lowest region of the atmosphere