Do you expect Delta H (the change in enthalpy) to be positive or negative for th

Question

Do you expect Delta H (the change in enthalpy) to be positive or negative for the reaction involving the formation of the aluminum/xylenol orange complex? Briefly explain your answer. Do you expect Delta S (the change in entropy) to be positive or negative for the reaction involving the formation of the aluminum/xylenol orange complex? Briefly explain your answer. What wavelength range do you expect lambda_max of the complex to fall in? Give a numerical answer. You mix together 3 mL of Al^3+ (4.00 times 10^-5 M) and 3 mL of xylenol orange (4.00 times 10^-5 M). The absorbance at room temperature is 0.1 and the absorbance at 65 degree C is 0.21. What is the concentration of AIQ at 65 degree C? K_eq = 3.46 times 10^-4 at 70 degree C. What is Delta G degree?

Explanation / Answer

Ans (1) n (2) is The reaction of aluminium ions n xylenol orange is

Al^3+ + H4Q --------? AlQ^- + 4H^+

Al. Ions xylenol complex

(Colorless) (yellow) (red)

The equilibrium constant K for the above reaction is

K = (AlQ^-) (H+)^4 / (H4Q) (Al^3+)

We can calculate K at several different temperatures.

Delta(G) = -RT ln K

Finally we can use the definition of gibbs energy.

Delta(G) = delta(H) - T(delta S)

A plot of delta(G) as a function of temperature gives a straight line with slope = -delta (S) and y intercept = delta(H).

Hence we expect change in enthalp i.e. delta (H) to be postive n change in entropy i.e. delta(S) to be negative for the reaction involving the formation of aluminium/ xylenol orange complex.

Ans 3) As the complex AlQ^- is a coloured complex so it absorb the wavelength of visible light . We expect the wavelength to be 550 nm bcoz at this wavelenghth the complex AlQ^- absorbs very strongly.

Ans 4) Absorbance at room temp = 0.1

Absorbance at 65 degree celsius= 0.21

For calculation, e2 - e1 = 2.50 × 10^4 where e2 n e1 are molar absorptivity

Formula for calculating concentration of AlQ^- at 65 degree celsius

= absorbance at 65 degree temp.- absorbance at room temp. / e2 - e1

= 0.21 - 0.1 / 2.50 × 10^4

= 0.14 / 2.50 × 10^4

= 0.056 × 10^-4 M

Ans 5) Given K eq. = 3.46 × 10^-4

Temperature = 70 degree celsius = 70 + 273 = 343 degree kelvin

Delta (G^¤ ) = - RT ( ln K eq.)

= - 2.303 RT ( log Keq.)

= - 2.303 × 8.314 × 343 ( log 3.46× 10^-4)

= -2.303× 8.314 × 343 × -3.4609

= 22729.36

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