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# The equilibrium ${{N}_{2}}+\text{ }{{\text{O}}_{2}}\text{ }\rightleftharpoons \text{ 2NO}$ is established in a vessel with a capacity of $2.5L$ . The initial amount of ${{N}_{_{2}}}$ is $2mol$ and that of ${{O}_{2}}$ is .Half a mole of ${{N}_{_{2}}}$ is used at the equilibrium. The molar concentration of $NO$ is:(A) $0.2$ (B) $0.4$ (C) $0.6$ (D) $0.1$

Last updated date: 14th Jul 2024
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Hint : We can solve this problem with the help of formula of molarity because the concentration in the problem is asked moles per litre so Molarity $=$ Number of moles $/$ Volume of solution and we know that number of moles $=$ mass of substance $/$ Molar mass of substance .Here molar mass of nitric acid and use the concept of initial concentration of reactant at time $\text{t}=0$ and the concentration of reactant and product at time $t={{t}_{eq}}$ and solve it for its equation.

Molar concentration can also be called molarity. The molar concentration can be defined as the measure of the concentration of the chemical species, in particular of the solute in the terms of amount of the substance per unit volume of solution. The most commonly used unit for molar concentration is moles per liter. Firstly let us look at the reaction which is:
${{N}_{2}}+\text{ }{{\text{O}}_{2}}\text{ }\rightleftharpoons \text{ 2NO}$
Now we know that when time is zero that means $\text{t}=0$ the concentration of the product will be zero as the reaction hasn’t started reacting yet and the amount of reactant is unused at that time. After some time of starting the reaction let us suppose that the reaction is in equilibrium now and it took time ${{t}_{eq}}$ to reach the equilibrium and we will write it as follows;
 ${{N}_{2}}$ ${{\text{O}}_{2}}$ $\text{ 2NO}$ At time $\text{t}=0$ the conc $2mol$ $4mol$ $0mol$ At time $t={{t}_{eq}}$ the conc. $2-x$ $4-x$ $2x$

Now we have written the equation for reaction at time $t={{t}_{eq}}$ and we are given that at time $t={{t}_{eq}}$ the used amount of reactant is $0.5\text{ mol}$ of ${{\text{N}}_{2}}$ the value of $x\text{ is 0}\text{.5 }$
So solving this for $x=0.5$ ; By Putting $x=0.5$
 ${{N}_{2}}$ ${{\text{O}}_{2}}$ $\text{ 2NO}$ At time $\text{t}=0$ the conc. By putting $x=0.5$ $2-0.5$ $4-0.5$ $2\times 0.5$ At time $t={{t}_{eq}}$ the conc. The following values $1.5$ $3.5$ $1$

Now we have found the value of moles of $NO$ at the time of equilibrium. Now we will use this value to find the molar concentration of $NO$ at the time $t={{t}_{eq}}$
So formula of molar concentration is = $\dfrac{m\text{oles of NO}}{Volume}$
Now the value of moles of $NO$ is $1$
And the value of Volume which is given in question is $2.5$
Putting values in formula we get m = $\dfrac{1}{2.5}$
Solving it we get the molar concentration = $0.4$
Therefore the correct answer is option B.

Note :
Remember that this type of question are solved using a pattern like firstly we see the initial concentrations when the reaction hasn’t started and then after reaction has started we know that some concentration of reactant will converted into product we say it is x amount and minus it from the reactant and the product formed will the multiple of coefficient of product and x and after that we will put the values given in question and solve it using the formula of molar concentration.