Calculating volume % in a mixture of gases at equilibrium - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-08-20T00:53:21Z https://chemistry.stackexchange.com/feeds/question/72799 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://chemistry.stackexchange.com/q/72799 0 Calculating volume % in a mixture of gases at equilibrium Freelancer https://chemistry.stackexchange.com/users/34526 2017-04-18T03:12:44Z 2018-02-24T16:49:18Z <blockquote> <p>In a mixture of $\ce{N2}$ and $\ce{H2}$ initially In a mole ratio of $1:3$ at $\ce{30 atm}$ and $\pu{300^\circ C}$, the percentage of ammonia by volume under the equilibrium is $17.8$. I have to calculate the equilibrium constant $K_\mathrm p$ of the mixture for the reaction $$\ce{N2(g) + 3H2(g)&lt;=&gt;2NH3(g)}$$</p> </blockquote> <p>When I read the question, the first thing that comes to me is Dalton's law of partial pressure and (I think) that according to that law we can imagine the components of gaseous mixture occupying a volume ($V$) each equal to the total volume of container. Thus, that is why we simply write the partial pressure of any gas as ($P_iV=n_iRT$) and sum all the partial pressure to get total pressure.</p> <p>Coming back to original question, according to the previous logic, each gas in the container at equilibrium is occupying same volume ($V$) equal to the total volume of the container. Then the term volume % doesn't make any sense because each gas is occupying the whole volume of the container (under ideal gas assumptions) so volume% of each gas $= 100\%$.</p> <p>So what does the question mean by "Volume %" (for gases)? Is there any flaw in the above logic?</p> <p>P.S: I would have easily solved the question if mole % was given as each gas has different moles so no problem there.</p>