How do you calculate the molar enthalpy of condensation ( $DeltaH_(cond)$ ) for ammonia when 100.0 g of $NH_3$ gas turn into a liquid at its boiling point?

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68,500 J of energy are released in the process.

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Answer 1 · 2017-06-19T17:19:15

By calculating the quotient, $"energy released"/"moles of ammonia"$ ......and get $DeltaH_"condensation"=-11.7*kJ*mol^-1$

We interrogate the process...........

$NH_3(g) rarr NH_3(l)+Delta_"condensation"$

We want $DeltaH_"condensation"$ with units of $kJ*mol^-1$ .

We have parameters for a $100*g$ mass of ammonia.....

And thus, $(-68500*J)/((100.0*g)/(17.03*g*mol^-1))=-11665.6*J*mol^-1$ , i.e. $-11.7*kJ*mol^-1$ .