Chemistry: Doubling Ammonia Concentration in 1L Dish

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To double the concentration of ammonia in a 1-liter dish containing 0.3 moles of nitrogen, 0.4 moles of hydrogen, and 0.1 moles of ammonia, the current concentrations are calculated as 0.3 M for nitrogen, 0.4 M for hydrogen, and 0.1 M for ammonia. Doubling the ammonia concentration requires reaching 0.2 M, which means increasing the moles of ammonia to 0.2 moles. The discussion focuses on determining how many additional moles of hydrogen are needed to achieve this new equilibrium while keeping the temperature constant. Participants express confusion about how to set up the reaction equation and calculate the necessary changes. The conversation emphasizes the need for clarity in understanding chemical equilibria and concentration adjustments.
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In a dish with the volume 1 litre we have 0.3 moles of nitrogen, 0.4 moles of hydrogen and 0.1 moles of amoniac. How many moles of hydrogen should we add to the dish so we can double the concentration of the balance for amoniac,if the temperature does not change?
So I find the concentrations by the formula C=n/V and they are respectively 0.3 0.4 0.1..now I don't know how to put the equation in the state of change...
 
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No idea what

Elaia06 said:
the concentration of the balance

is.
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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