A student wants to prepare a 1.0-liter solution of a specific Molarity. The student determines that the mass of the solute needs
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<u>Answer:</u> The equilibrium concentration of NO after it is re-established is 0.55 M
<u>Explanation:</u>
For the given chemical equation:
The expression of for above equation follows:
.....(1)
We are given:
Putting values in expression 1, we get:
Now, the concentration of NO is added and is made to 0.700 M
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
The equilibrium will shift in backward direction.
<u>Initial:</u> 0.200 0.200 0.700
<u>At eqllm:</u> 0.200+x 0.200+x 0.700-2x
Putting values in expression 1, we get:
So, equilibrium concentration of NO after it is re-established = (0.700 - 2x) = [0.700 - 2(0.075)] = 0.55 M
Hence, the equilibrium concentration of NO after it is re-established is 0.55 M
Explanation:
When work is done then there will occur change in volume. And, most change in volume occurs when there will be production of gas is taking place. We assume that no work is done when no gas is produced.
(a) For
Here, 3 moles of gas is producing 0 moles of gas. This means that work is done on the system.
(b) And, more is the production of a gas taking place in a reaction more will be the amount of work done by the system.
For
Here, 2 moles of a gas is producing 3 moles of a gas. Since, gas is increasing so, work will be done by the system.