Please explain and provide an answer to the question:
Nitrogen Gas (N2) and Hydrogen Gas (H2) combine to form Ammonia (NH3) at a ratio of 1:3:2
This is a reversible chemical equation.
N2 + 3H2 <-> 2NH3
An unknown amount of Ammonia was added to a 1.0 L container. When the reaction reached its equilibrium, the concentration of Nitrogen gas was found to be 2.00 mol/L . How many moles of Ammonia were added if the Equilibrial Constant is 2.0?
I'm contemplating whether this will require the application of Le Chatelier's Principle.. So im not sure about the answer...
Nitrogen Gas (N2) and Hydrogen Gas (H2) combine to form Ammonia (NH3) at a ratio of 1:3:2
This is a reversible chemical equation.
N2 + 3H2 <-> 2NH3
An unknown amount of Ammonia was added to a 1.0 L container. When the reaction reached its equilibrium, the concentration of Nitrogen gas was found to be 2.00 mol/L . How many moles of Ammonia were added if the Equilibrial Constant is 2.0?
I'm contemplating whether this will require the application of Le Chatelier's Principle.. So im not sure about the answer...
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Eq. const. K = [NH3]^2 / [N2] [H2]^3
1 mole NH3 when dissociated forms 1/2 mole N2 and 3/2 moles of H2
Since N2 formed was 2 mole then H2 formed must be 6 moles and NH3 decomposed must be 4 moles
Suppose initially 'x ' mole NH3 was added then
K = [x - 4]^2 / [2] [6]^3 =2 when 'x' = 25.4
If the eq. const. K is for reverse reaction then K = [2] [6]^3 / [x -4]^2 =2 and 'x' = 10.4
1 mole NH3 when dissociated forms 1/2 mole N2 and 3/2 moles of H2
Since N2 formed was 2 mole then H2 formed must be 6 moles and NH3 decomposed must be 4 moles
Suppose initially 'x ' mole NH3 was added then
K = [x - 4]^2 / [2] [6]^3 =2 when 'x' = 25.4
If the eq. const. K is for reverse reaction then K = [2] [6]^3 / [x -4]^2 =2 and 'x' = 10.4
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Unlike a math equation, no chemical equation can be reversed, or it will not work.