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[VISUAL] In an experiment to measure the strength of the intramolecular hydrogen-bond in B, the chemPhysical Chemistry Chemistry Question

Hydrogen bond strength determination

[VISUAL]

In an experiment to measure the strength of the intramolecular hydrogen-bond in B, the chemical shift of the amide proton δobs, was measured at various temperatures.

The observed chemical shift, δobs, is the weighted average of the shifts of the N–H proton when the amide is completely hydrogen bonded, δh, and when it is completely free, δf.

Data Table:
| T / K | δobs / ppm |
|---|---|
| 220 | 6.67 |
| 240 | 6.50 |
| 260 | 6.37 |
| 280 | 6.27 |
| 300 | 6.19 |

17.1.

Derive an expression for the observed chemical shift of the N–H proton, δobs.

Model Answer

δobs = xh*δh + xf*δf
where xh and xf are the mole fractions of the hydrogen bonded species and the free species, respectively, and so xh + xf = 1.

17.2.

Derive an expression for the equilibrium constant K for A ⇌ B in terms of δobs, δh, and δf.

Model Answer

K = xh/xf

δobs = xh*δh + xf*δf = xh*δh + (1 − xh)*δf ⇒ xh*(δh − δf) = δobs − δf
also
δobs = (1 − xf)*δh + xf*δf ⇒ xf*(δf − δh) = δobs − δh

K = xh/xf = (δobs − δf) / (δh − δobs)

17.3.

Given that δh = 8.4 ppm and δf = 5.7 ppm, calculate the equilibrium constants for the cyclisation at the different temperatures.

Model Answer

T / K | δobs | K
220 | 6.67 | 0.5607
240 | 6.5 | 0.4211
260 | 6.37 | 0.3300
280 | 6.27 | 0.2676
300 | 6.19 | 0.2217

17.4.

By plotting a suitable graph, determine the standard enthalpy change for A → B and the standard change in entropy at 300 K.

Model Answer

A plot of lnK vs 1/T gives a straight line with slope (= – ∆rH°/ R) = 764.1 K .

∆rH° = –764.1 × 8.3145 J mol–1 = – 6.4 kJ mol–1
∆rS°(300) = ((∆rH° – ∆rG°(300)) / 300) J K–1 mol–1 = – 34 J K–1 mol–1

17.5.

Discuss the significance of your answers to part (17.2).

Model Answer

The enthalpy change is exothermic, which is not surprising since a new bond is formed. However, the value is much smaller than that for forming a full covalent bond.

The entropy change is negative due to the loss of rotational freedom as the chain becomes a ring.

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