10) D) 10.60 1o) At what pH is the amino acid glycine with a Ka of 251- 10-10 si
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10) D) 10.60 1o) At what pH is the amino acid glycine with a Ka of 251- 10-10 sinty-stx (760) percent dissociated? 10.10 A) 9.60 ) 9.89 11) The chemical system shown below is at equilibrium Which change is at equilibrium. Which change in conditions will not result in 11) s forward reaction? B) adding more N2 D) adding a catalyst A) reducing the volume adding more H2 12) For which process is the sign of AS negative in the system? A) H2O(0) H2Og) B) 2 H2(8).O2) 2H20(g) C) H20(0)-H2O) D)2H20() + 2 K(s)--2 K.(aq) .2 OH-(aq) + H2(g) 13) Sodium reacts violently with water according to the equation: 13) 2 Na(s) 2 H2o(0)-2 NaOH(ag) H2G) The resulting solution has a higher temperature than the water prior to the addition of sodium. What are the signs of ??? and ASe for this reaction? A) AH is positive and AS is positive. C) AH is negative and AS is positive. B) dF is positive and AS is negative. DAH' is negative ard AS is negative. 14) Predict the sign of ?S of the system for both of the following. 14) I. 11. 2 Cgraphite) + O2 (g)-2 doe) C4H10 ()--C4H10 (g) A) AS should be positive for I and negative for II. B) AS should be negative for I and positive for II. C) S should be positive for 1 and positive for D) AS should be negative for 1 and negative for ?. 15) For which of the following will the entropy of the system increase? 15) A) reaction of magnesium with oxygen to form magnesium oxide B) reaction of nitrogen and hydrogen to form ammonia C) condensation of steam D) sublimation of ice 16) Which has the highest standard molar entropy at 25 C? 16) A) Al (s) B) Al () C Al (g) D) All three should have a standard molar entropy of zero. 17) For the reaction 3 C2H2(g) CH6) at 25 C, the standard enthalpy change is -631 kJ and the 17) standard entropy change is-430 J/K. Calculate the standard free energy change at 25°C A) -618 k B)-503 k C-1061 k D) 948 kExplanation / Answer
10) seventy six percent dissociation means, the concentration of salt = 0.76 X Initial concentration of acid
final concentration of acid = 0.24 X Initial concentration
Let iniital concentration = M
The pH will be calculated as
pH = pKa + log [salt] / [acid]
pKa = -logKa= 9.6
pH = 9.6 + log 0.76M / 0.24 M = 9.6 + 0.5 = 10.10
11) The reaction will go in forward direction
On adding reactant = the addition of H2 or N2
If we reduced volume the reaction will go in the direction where the number of moles are less. again forward direction
Although the answer is simple: Catalyst shows no effect on equilibrium.
12) The sign of DeltaS will be negative for the reaction where less number of gaseous products are present
Answer: 2H2(g) + O2(g)--->2H2O(g)
13) The reaction is spontaneous,
T is higher after reaction means the reaction is exothermic, hence DeltaH < 0 or negative
More gaseous molecules on product side so the entropy increases hence Delta S > 0 or positive
14) More gaseous products in both the reactions so Delta S will be positive for both the reactants
15) Sublimation of ice will increase the entropy
16) Al(g)
17) DeltaG = DeltaH- TDeltaS
DeltaG = -631kJ - (298.15 X (-0.430 kJ / K) = -502.79 kJ = -503 kJ
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