Consider the following reaction equilibrium N2(g)+3H2(.g.)leftharpoons2NH3(.g.) Initially, 1 mole of N2 and 3 moles of H2 are taken in a 2 L flask. At equilibrium state if, the number of moles of N2 is 0.6, what is the total number of moles of all gases present in the flask?

Question

Consider the following reaction equilibrium N2(g)+3H2(.g.)leftharpoons2NH3(.g.) Initially, 1 mole of N2 and 3 moles of H2 are taken in a 2 L flask. At equilibrium state if, the number of moles of N2 is 0.6, what is the total number of moles of all gases present in the flask? (A) 0.8 (B) 1.6 (C) 3.2 (D) 6.4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-4hfj5hvvhwwi.png" /> Number of moles of textN2=0.6=1- textx ∴ textx=1-0.6=0.4 So, 3-3 textx=3-3× 0.4=1.8 , and 2 textx=2× 0.4=0.8 Therefore, the total number of moles at equilibrium. =(1 - textx)+(3 - 3 textx)+2 textx =0.6+1.8+0.8 =3.2 mol

Q. Consider the following reaction equilibrium N2​(g)+3H2​(g)⇌2NH3​(g) Initially, 1 mole of N2​ and 3 moles of H2​ are taken in a 2 L flask. At equilibrium state if, the number of moles of N2​ is 0.6, what is the total number of moles of all gases present in the flask?

0.8

1.6

3.2

6.4

Number of moles of N2​=0.6=1−x ∴ x=1−0.6=0.4 So, 3−3x=3−3×0.4=1.8 , and 2x=2×0.4=0.8 Therefore, the total number of moles at equilibrium. =(1−x)+(3−3x)+2x =0.6+1.8+0.8 =3.2 mol

Number of moles of $N_{2} = 0.6 = 1 - x$

$∴$ $x = 1 - 0.6 = 0.4$

So, $3 - 3 x = 3 - 3 \times 0.4 = 1.8$ ,

and $2 x = 2 \times 0.4 = 0.8$

Therefore, the total number of moles at equilibrium.

$= (1 - x) + (3 - 3 x) + 2 x$

$= 0.6 + 1.8 + 0.8$

$= 3.2$ mol