Question

What is approximate value of log \[{K_p}\] ​ for the reaction: \[{N_2}\left( g \right) + 3{H_2}\left( g \right) \leftrightharpoons 2N{H_3}\left( g \right)\;\] at \[{25^0}C\]? The standard enthalpy of formation of \[N{H_3}\left( g \right)\;\] is \[ - 40.0\;kJ/mol\;\] and standard entropies of \[{N_2}\left( g \right),{H_2}\left( g \right)\;\] and \[N{H_3}\left( g \right)\;\] are \[191,130\;\] and \[192\;J{K^{ - 1}}mo{l^{ - 1}}\] respectively.
(A) $0.04$
(B) $7.05$
(C) $8.6$
(D) $3.73$

Answer 1

Hint: To solve this, we have to calculate the total enthalpy change in the reaction, and standard entropy change then we will use the Gibbs free energy equation to calculate the value of log \[{K_p}\] with the help of given values.

FORMULAE USED:
The standard change in Gibbs free energy is given by the equation:
\[\Delta G = \Delta H - T\Delta S\]
It is also given by the equation:
\[ - \Delta G = 2.303RT\;log{K_p}\]

Complete step-by-step answer:We are given the equation:
\[{N_2}\left( g \right) + 3{H_2}\left( g \right) \leftrightharpoons 2N{H_3}\left( g \right)\;\]
Also, the given standard enthalpy of formation of \[N{H_3}\left( g \right)\;\]$ = - 40.0\;kJ/mol\;$
Value of standard entropy of \[{N_2}\left( g \right) = 191J/Kmo{l^{ - 1}}\]
Value of standard entropy of \[{H_2}\left( g \right) = 130\,J/Kmo{l^{ - 1}}\]
Value of standard entropy of \[N{H_3}\left( g \right)\; = 192\,J/Kmo{l^{ - 1}}\]
Now, we will calculate the net enthalpy of formation from the equation, it will be:
\[\Delta H = - 40 \times 2 = - 80.0\;\,kJ\]
Now, we will calculate the net change of entropy from the equation, it will be:
\[\Delta S = 2 \times {S_{N{H_3}}} - {S_{{N_2}}} - 3 \times {S_{{H_2}}}\]
\[\Delta S = 2 \times 192 - 191 - 3 \times 130 = - 197J\]
Now, we will calculate the change in Gibbs energy from the equation of Gibbs energy, it will be:
\[\Delta G = \Delta H - T\Delta S\]
Putting the value of \[\Delta H\]and \[\Delta S\] temperature in the equation we get;
\[\Delta G = - 80 \times {10^3} - 298 \times ( - 197) = - 21294J\]
Now, we will put this value in the second equation of Gibbs free energy to calculate the value of log \[{K_p}\] ​ , it will be:
\[ - \Delta G = 2.303RT\;log{K_p}\]
\[21294 = 298 \times 8.314 \times \;2.303\,log{K_p}\]
\[log\;{K_p} = 3.73\]

Hence, the value of log \[{K_p}\] ​ for the given reaction will be $3.73$. Therefore, option (D) is correct.

Note: Gibbs free energy is a thermodynamic quantity that is used to measure the maximum amount of work done in a thermodynamic system at a constant value of temperature and pressure. It is denoted by the symbol ‘G’ and is generally expressed in Joules or Kilojoules. It can also be expressed as the maximum amount of work that can be done from a closed system.