Answer to Question #119135 in General Chemistry for Abigail Rouse

Question

A system at equilibrium is represented by the following chemical equation: N₂(g) + 3H₂(g) <--> 2NH₃(g) At a given temperature, 1.20 moles of N₂, 0.866 moles of H₂, and 4.17 * 10⁻⁸ moles of NH₃ are in the 4.00 L vessel that contains the equilibrium. Calculate the value of K for the system.

Answer 1

Solution.

$n(N_2)=1.20mol;$

$n(H_2)=0.866mol;$

$n(NH_3)=4.17\sdot10^{-8}mol;$

$V=4.00L;$

$N_2+3H_2\leftrightarrow 2NH_3;$ $K=\dfrac{[NH_3]^2}{[N_2][H_2]^3};$

$[NH_3]=\dfrac{n(NH_3)}{V};$

$[NH_3]=\dfrac{4.17\sdot10^{-8}mol}{4.00L}=1.0425\sdot10^{-8}M;$

$[N_2]=\dfrac{n(N_2)}{V};$

$[N_2]=\dfrac{1.20mol}{4.00L}=0.3M;$

$[H_2]=\dfrac{n(H_2)}{V};$

$[H_2]=\dfrac{0.866mol}{4.00L}=0.2165M;$

$K=\dfrac{(1.0425\sdot10^{-8}M)^2}{0.3M\sdot(0.2165M)^3}=1.673\sdot10^{-15}$;

Answer: $K=1.673\sdot10^{-15}.$