Achieve week 1 #6

[Chloe Fuson]

I was wondering if someone could explain how to do the following problem. I know that the correct answer is Q= 0.061 (I got there through process of elimination), but for some reason when I do my calculation using the expression for the reaction quotient I am not coming up with that answer, so I just want to make sure I'm not missing a step.

6) Consider the reaction at 500°C .

N2(g)+3H2(g)↽−−⇀2NH3(g) Kc=0.061

If analysis shows that the composition of the reaction mixture at 500°C is 1.14 mol⋅ L−1N2 , 5.52 mol⋅ L−1H2 , and 3.42 mol⋅ L−1NH3 , what is the value of the reaction quotient Q ?

[carterwink2K]

(3.42^2)/((1.14)(5.52^3)) = 0.061

[Rhea Desai 1A]

The reaction quotient has the same equation as K so Q = [NH3]^2 / [N2][H2]^3 = [3.42]^2 / ([1.14] * [5.52]^3) = 11.7/ (1.14 * 168.2) = 11.7 / 191.74 = 0.061
hope this helps

[Gabrielle Malte 2G]

Since all molecules are in the gas phase, they are all included in the equations used to calculate Q and K. Therefore, the equation for Q should be:

Q= [NH3]^2/([H2]^3*[N2]) = (3.42)^2/((5.52)^3*(1.14)) = 0.061

[trevina_brown_2A]

The reaction quotient has the same equation as K so Q = [NH3]^2 / [N2][H2]^3 = [3.42]^2 / ([1.14] * [5.52]^3) = 11.7/ (1.14 * 168.2) = 11.7 / 191.74 = 0.061
hope this helps

This was helpful! Also the feedback on achieve is very helpful as well

[Mrinalini Mishra 2L]

Since all molecules are in the gas phase, they are all included in the equations used to calculate Q and K. Therefore, the equation for Q should be:

Q= [NH3]^2/([H2]^3*[N2]) = (3.42)^2/((5.52)^3*(1.14)) = 0.061

I ha questions about this same problem, but you explained it really well here. Thank you!

[Allen Lu 2F]

The reaction quotient is calculated the same way as the equilibrium constant. In the case of this problem, you use the values given to calculate Q: [products]/[reactants]. The result will yield Q=0.061 which is the same as Kc.

[Jonathan Alterman 1C]

You calculate the reaction quotient Q the same way as the equilibrium constant K. Therefore, Q would equal concentration of products over concentration of reactants each to the power of their respective stoichiometric coefficients.

[Adithi Ayyala 2G]

Yep, we can solve for Q the same way we would for K, and then substitute the values into the equation. Finally, we can compare Q and K values to understand the motion of the equation (favoring the reactants versus products). When Q>K, the reaction is exothermic (favors forward rxn), whereas is Q<K, the reaction is endothermic (favors reverse rxn), and when the two are equal, the system is at equilibrium.

[Vashe Sundar 3H]

Q can be calculated the same way you find K. So input the end concentrations into: [NH3]^2/[N2][H2]^3

[Riya Sawhney 1C]

Here is how I did it!

Q=[NH3]^2/([N2][H2]^3)= (3.42)^2/(1.14)(5.52)^3 =0/.061 (using current composition of mixture)

Since Kc=Q we can conclude that the reaction is already at equilibrium.