uc Davis CHE 2B win x For many weak acid or w x C For Many weak Acid or x G When
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uc Davis CHE 2B win x For many weak acid or w x C For Many weak Acid or x G When calcul the equ x C www.sap ilglea Vibisc od/ibis/view.php?id-3259156 Apps D hde mgmt-026-chapte e Chemicals used on Go RCEr Education-CIIE RECI Education Couu D D ® Mini school bus I landicap bus unfi ion Use and Misuse of l is Print II Calculator Herhodic able Available From Question 29 uf 33 Inuurruct ncorruct ncurroc r1uurruct 00 Due Date: Sapling Learning Points Possible 20 Grade Categor 00 When determining the equilibrium concentrations for the reaction Description A B 95 22 AB (g, Policies: AB 00 23 You can check a simplifying assumption can be used under certain conditions to avoid solving a quadratic equation. 00 You can view s up on any ques AB -x A, B 25 95 You can keep t you get it right Classify each situation by whether the simplifying assumption can be used or whether the quadratic formula 26 95 is required. You lose 5% of Use simplifying assumption Quadratic formula required 27 00 n your questic [AB 0.00243 M (AB 0.339 M ansWel Ko 89 x 10e Ko 0.00230 28 00 AB 0.0362 M Kc 3.84 x 10e Textbook 29 AB 0.0358 M Ko 0.00011 o Help with Th 30 00 AB 0.389 M Ke 0.000693 Web Help & 32 00 Technical Sul O Previous Give Up & View solution Try Again Next d Exit Explanation nn 22 2.06 AM P a 2/16/2017Explanation / Answer
The given equation is
AB (g) <=====> A (g) + B (g)
Kc = [A][B]/[AB]
The simplified equilibrium constant may be written as
Kc = x2/([AB] – x) x2/[AB]
1) [AB] = 0.00243 M, Kc = 1.89*10-6
We can ignore x in the denominator when Kc is extremely low. The logic remains that since Kc is small, the reaction essentially favours the reactant side (reverse reaction). Hence x can be ignored in the expression.
2) [AB] = 0.339 M, Kc = 0.00230
Kc is small, but definitely not as small as in (1) above. The initial reactant concentration is appreciably high (much higher than in 1 above) and the extent of the reaction (x) is quite small. We can therefore, ignore x in the denominator.
3) [AB] = 0.0362 M, Kc = 3.84*106
Kc is large, positive. The reaction essentially favours the product side (forward reaction). Hence, we cannot assume that whatever concentration of the reactant we started with, we will be left with the same reactant concentration at equilibrium. Therefore, we cannot ignore x here and we need to solve the quadratic equation.
4) [AB] = 0.0358 M, Kc = 0.000111
Kc is small, but not as small as in (1) above. However, the initial reactant concentration is also small. If we ignore x here, will assume that the reactant concentration (small itself) remains unchanged which is not true. Hence, we shouldn’t ignore x here.
5) [AB] = 0.389 M, Kc = 0.000693
Kc is small, but the initial reactant concentration is appreciably high. Hence, we can ignore x here.
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