Question

CHEMICAL EQUILIBRIUM!!
Consider the following reversible reaction. In a 3.00-liter container, the following amounts are found in equilibrium at 400°C: 0.0420 mole N2, 0.516 mole H2, and 0.0357 mole NH3. Evaluate
Kc
. N2(g) + 3H2(g) <---> 2NH3(g)
a. 0.202 b. 1.99 c. 16.0 d. 4.94 e. 0.503

Answer is B
I understand the setup but what do I do with the 3.00L and 400 degrees C? How could I get the answer b??

Answer 1

The equilibrium constant at 400 degrees C for the reaction:
\[\large Kc _{T=400}=\frac{ \left[ NH _{3}(g) \right]^{2} }{ \left[ N _{2}(g) \right] \left[ H _{2}(g) \right]^{3} }\]
The concentrations are defined as:
\[\large C=\frac{ n }{ V }\]

Answer 2

Substitute and we get the following:
\[\LARGE Kc _{T=400}=\frac{ \left( \frac{ n(NH _{3}(g)) }{ 3.00l } \right)^{2} }{ \frac{ n(N _{2}(g)) }{ 3.00l } \left( \frac{ n(H _{2}(g)) }{ 3.00l } \right)^{3}}\]

Answer 3

For evaluation see attachment.

Answer 4

All the measurements have been done at 400 degrees C therefor the equilibrium constant is also at 400 degrees C.

Answer 5

When calculating the answer the way it was set up in your attachment the answer comes out to 1.987 x 10 ^-10. The answer you gave was just the 1.987 which would correspond to the supposed correct answer as it is given in option b which is 1.99. I'm confused how I got the answer but with a power of -10?

Answer 6

Try look if your brackets are set correctly else try calculate it in parts. Or you can take a screenshot if using a calculator on your PC and let me go through the calculation.