AGAH-TAS = - RT⁰ In K AGR In K = - rot+ AHº RT AS⁰ R (1) (2) (3) Show, without leaving out any steps, the derivation of Eqn. (3) from Eqns. (1) and (2).
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![AG=AH-TAS
AGR
=-RT° In K
ΔΗ
AS⁰
In K = - rot+
sp
RT
R
(1)
(2)
(3)
Show, without leaving out any steps, the derivation of Eqn. (3) from Eqns. (1) and (2).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3192a6a6-9924-4cea-a59f-2dfeefe422a5%2F185bac74-7d9f-4592-905a-f06a0a3956ca%2Fzk7jbbm_processed.jpeg&w=3840&q=75)
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- A 25.00cm of potassium dichromate(K,Cr,CH so lution aretadded to an acidified aqueous solution of potassium lodide(present in excess). The lodine was then liberated according to the equation K-Cr.O;(aq) +14- (aq) + KI(ag) K(aq) + Cr (aq) + H;O(1) + 12(aq) Required 19.63cm of a 0.1478M Na S,O for complete reaction. The equation for the reaction of between Na S;03 and I2 is 2(aq)+ Na;S;SO;(aq)-Na S4O6(aq) + Nallac) Calculate the concentration of potassium dichromate (only 4 decimal places) mol/LDifferentiate between circumstances for the use of nRln(V2/V1) ??? Cvln(T2/T1) in calculating change in S7. Solve for T q = mc(T; – T;) тс |
- Part IV: Equilibrium Involving Chromate and Dichromate Ions Steps 2 & 3 Step Steps 5 & 6 4 Steps 7 & 8 Step 9 2CrO₂² (aq) + 2 H (aq) Cr₂O, (aq) + H₂O (1) Initial Colour + 1 M NaOH + 1 M HC1 Ba² (aq) + CrO² (aq) BaCrO. (s) 0.1 MK₂CrO₂ Yellow Yellow Yellow Initial Colour + 1 M HC1 + 1 M NaOH 0.1 MK₂CrO₂ + 0.1 M Ba(NO3)2 Other Observations: 0.1 MK₂CrO₂ Yellow Orange Yellow Initial Colour Yellow 0.1 MK₂Cr₂O, Orange Initial Colour + 1 M NaOH Yellow + 1 M HC1 Orange 0.1 MK₂Cr₂O, Orange Yellow Orange 0.1 MK₂Cr₂O₂ + 0.1 M Ba(NO3)2 Orange Orange Orange Immediately cloudy with white precipitate and yellow solution + 0.1 M Ba(NO₂)₂ Orange + 1 M HC1 Orange with precipitation + 1 M NaOH Yellow with precipitation 0.1 MK₂CrO₂ 0.1 MK₂Cr₂O, Immediately a lot of precipitate Very little precipitate formed in formed in yellow solution orange solutiondard hee en brus olooj of eolrojolix > 4CO2 + 6H,0 2. Calculate AG° for the combustion of ethane 2C,H6) + 702) (8)9- 28) 2AT with the following AG°;: CH = -32.86 kJ/mol 5 1 Rrogr CO2 =-394.4 kJ/mol orle ridw svilieoq ei vgron 1ovisa H,O = -237.2 kJ/mol plpnei uovitieoq o16 amnsh (),16.72 Show how P = kT(a In Q/OV), leads to PV = nRT.
- distinguish between the circumstances for the use of nRln(V2/V1) and Cvln(T2/T1) in calculating change in Stes (18) ades Chemistry: Q4: 2022-23 | e 4.4Be + 2He→ 6C+ Ⓒin O On O in 9 Sie le(4.113+1 5. A monoprotic acid (HA) with K, of 8.76 x 10 has a partition coefficient of 3.9 (favoring octanol) when distributed between water and octanol, An aqueous solution (100.0 ml, pH 4.50) initially containing a 0.200 M formal concentration of HA is extracted once using 30.0 mL of octanol. a) What is the extraction efficiency? b) Calculate the formal concentration of the acid in both the water phase and the octanol phase after performing the extraction. Fwater = Foctanol = c) Would the extraction efficiency increase, decrease, or not change if the pH of the aqueous phase is adjusted to 7.00 prior to performing the extraction? no change Circle one: increase decrease Briefly explain:
- 1. (a) Calculate Eo’ for the process:AgI03(s) + e- ⇔ Ag(s) + I0 3-(b) Use the shorthand notation to describe a cell consisting of a saturated calomel reference electrodeand a silver indicator electrode that could be used to measure pIO3.(c) Develop an equation that relates the potential of the cell in (b) to pIO3.(d) Calculate plO3 if the cell in (b) has a potential of 0.294 V.8.8 The e.m.f. of the cells Mo „Mo02|(Zro2) (Ca0)o.15|Fe,Fe0, 0.85 Mo „Mo02|(Zr02), and (Cao)o.15|Ni,Nio 0.85 at 800°C (1073 K) is 17.8 and 284.0 mV respectively. Calculate the standard free energy change for the reaction Fe0 + Ni Ni0 + Fe. Ans 12,274 cal.6D.4 The potential of the cell Pt(s)|H,(g,p®)[HCI(aq,b)|Hg,Cl,(s)|Hg(1) has been measured with high precision with the following results at 25°C: b/(mmolkg") 1.6077 5.0403 3.0769 7.6938 10.9474 E/V 0.600 80 0.568 25 0.543 66 0.522 67 0.505 32 Determine the standard cell potential and the mean activity coefficient of HCI at these molalities. (Make a least-squares fit of the data to the best straight line.)
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