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Ammonia (NH3) is the active cleaning ingredient in Windex and is also the main contributor to the odor of stale cat urine. Ammonia has a ΔH°vap of 23.35 kJ/mol and a ΔS°vap of 97.43 J/mol·K. What is the normal boiling point of ammonia

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sebassandin

Answer:

[tex]T_b=239.7K=-33.49\°C[/tex]

Explanation:

Hello!

In this case, since the relationship between entropy and enthalpy for any process is defined below:

[tex]S=\frac{H}{T}[/tex]

For the vaporization of ammonia or any liquid, we can write:

[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T_{vap}}[/tex]

In such a way, solving the temperature of vaporization, or boiling point, we have:

[tex]T_{vap}=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]

Plugging in the given enthalpy and entropy of vaporization we obtain:

[tex]T_{vap}=T_b=\frac{23350\frac{J}{mol} }{97.43\frac{J}{mol*K}} \\\\T_b=239.7K=-33.49\°C[/tex]

Best regards!

pstnonsonjoku

From the information provided in the question, the boiling point of ammonia is 240 K.

Entropy is the degree of disorderliness in a system. The entropy of a system can be obtained using the relation;

ΔS°vap = ΔH°vap /T

Now;

ΔH°vap = 23.35 kJ/mol

ΔS°vap = 97.43 J/mol·K

T = ?

Making T the subject of the formula and substituting values;

T = ΔH°vap /ΔS°vap

T = 23.35 × 10^3J/mol/97.43 J/mol·K

T = 240 K

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