The reaction NO1g2 + NO21g2 ∆ N2O31g2 takes place in the atmosphe...

Question

The reaction NO1g2 + NO21g2 ∆ N2O31g2 takes place in the atmosphere with Kc = 13 at 298 K. A gas mixture is prepared with 2.0 mol NO and 3.0 mol NO2 and an initial total pressure of 1.65 atm.
(b) What is the volume of the container?

Accepted Answer

Welcome back everyone at 2 98 kelvin. The reaction where two moles of phosphorus tri chloride gas reacting with oxygen gas to produce two moles of phosphorus oxy chloride gas with an equilibrium constant of 16 occurs in the atmosphere with 2.4 moles of phosphorus tri chloride in 3.1 moles of oxygen and an initial total pressure of 1.38 A t M's. A gas mixture is created. What is the container's volume? So we want to recall our formula with which relates pressure times volume. Sorry, this should be pressure times volume set equal to our molds of our gas times. R gas constant R times our temperature in kelvin. And because our prompt asks us for the containers volume, we're isolating for volume. So that we can say that volume is equal to the most of our gas times. R gas constant. R times temperature divided by pressure, where we need to figure out what end our molds of gasses and that is why we can use our info from the prompt for each of our moles of gas to get the sum for our reactant here. And we would say that for our initial moles of gas From the prompt, we have 2.4 moles of our phosphorous tri chloride gas added to our 3.1 moles of our second reactant being oxygen gas and this some gives us our moles of gas being 5.5 moles initially. So we're going to use this for our formula and say that volume is equal to 5.5 moles of gas multiplied by r gas constant R Which we should recall a 0.8206 moles times. Leaders divided by, sorry, not most, but divided by a t M's times Calvin, Which is then multiplied by our temperature, which in the prompt we're told temperature in the proper units of Kelvin as 298 Kelvin. And then this is divided by our pressure, which in the prompt is given to us as 1.38 ATMs. And so we can cancel our units of a t M. S with a t M's in the denominator as well as we can get rid of moles and we can also get rid of kelvin leaving us with leaders as our final unit of volume. And so we can carefully plug this into our calculator and get a volume for the container equal to 97.4611 leaders which we can round two. Exactly 26 fix As just leaders. And this would be our final answer for the container's volume. Once our gas mixture is created. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video

Watch next