Equilibrium will shift towards the products when temperature is decreased in an exothermic reaction of the formation of ammonia.
<h3>What is an exothermic reaction?</h3>
An exothermic reaction is a reaction in which heat content of the reactants is greater than the heat content of product.
In an exothermic reaction, heat is given off.
For an exothermic reaction in equilibrium, increasing temperature shifts equilibrium to the towards the left, towards the reactants.
On the other, equilibrium will shift towards the products when temperature is decreased.
Therefore, equilibrium will shift towards the products when temperature is decreased in the reaction of the formation of ammonia.
Learn more about exothermic reactions at: brainly.com/question/13892884
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Answer:
It favors the forward reaction.
Explanation:
According to Le Chatelier's Principle, when a system at equilibrium suffers a perturbation, the system will react in order to counteract the effect of such perturbation.
If more reactant is added, the system will try to decrease its concentration. It will do so by favoring the forward reaction, decreasing the concentration of the reactant and increasing the concentration of the products, in order to re-establish the equilibrium.
Answer:
You will never know the exact volume with charles law
Explanation:
Doubling the temperature of gas doubles its volume, so long as the pressure and quantity of the gas are unchanged.
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
