Hey there !
Given the reaction:
N2 + 3 H2 = 2 NH3
At constant pressure and temperature ,volume is proporcional to moles:
Theoretical moles of N2 and H2 => 1:3
Theoretical volume of N2 and H2 => 1:3
Experimental volume of N2 and H2 => 3.0 L : 4.0 L
0.75 : 1 = 2.25 : 3
Since N2 is in excess reactant , H2 is the limiting reactant
Therefore:
volume of NH3 is 2/3 * Volume of H2
= 2/3 * 4.0 = 2.66 L
Hope that helps!
I would say C. (2,3). If you input the values into the equation, it works out. If you put 2 in as x and y as 3, it’s 2 + 1 = 3 which obviously works. Also, if you plot the point (2,3) on the graph, the line intersects it.
He bought 10 bottles a month at $1.00 and 8 bottles a month at $1.50.
This problem is asking for the equilibrium constant at two different temperatures by describing the chemical equilibrium between gaseous nitrogen, oxygen and nitrogen monoxide at 25 °C and 1496 °C as the room temperature and the typical temperature inside the cylinders of a car's engine respectively:
N₂(g) + O₂(g) ⇄ 2 NO(g)
Thus, the calculated equilibrium constants turned out to be 6.19x10⁻³¹ and 9.87x10⁻⁵ at the aforementioned temperatures, respectively, according to the following work:
There is a relationship between the Gibbs free energy, enthalpy and entropy of the reaction, which leads to the equilibrium constant as shown below:

Which means we can calculate the enthalpy and entropy of reaction and subsequently the Gibbs free energy and equilibrium constant. In such a way, we calculate these two as follows, according to the enthalpies of formation and standard entropies of N₂(g), O₂(g) and NO(g) since these are assumed constant along the temperature range:

Then, we calculate the Gibbs free energy of reaction at both 25 °C and 1496 °C:

And finally, the equilibrium constants derived from the general Gibbs equation and Gibbs free energies of reaction:
![K=exp(-\frac{\Delta _rG}{RT} )\\\\K_{25\°C}=exp[-\frac{172420 J/mol}{(8.3145\frac{J}{mol*K})(298.15K)} ]=6.19x10^{-31}\\\\K_{1496\°C}=exp[-\frac{135650J/mol}{(8.3145\frac{J}{mol*K})(1769K)} ]=9.87x10^{-5}](https://tex.z-dn.net/?f=K%3Dexp%28-%5Cfrac%7B%5CDelta%20_rG%7D%7BRT%7D%20%29%5C%5C%5C%5CK_%7B25%5C%C2%B0C%7D%3Dexp%5B-%5Cfrac%7B172420%20J%2Fmol%7D%7B%288.3145%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%28298.15K%29%7D%20%5D%3D6.19x10%5E%7B-31%7D%5C%5C%5C%5CK_%7B1496%5C%C2%B0C%7D%3Dexp%5B-%5Cfrac%7B135650J%2Fmol%7D%7B%288.3145%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%281769K%29%7D%20%5D%3D9.87x10%5E%7B-5%7D)
Learn more:
Liquids flow because its particles can move over each other. The particles have freedom to move and they will keep moving until it takes the shape of its container.
Solid particles are rigid and tightly packed, they cannot move, so it does not flow.