Nitrogen trifluoride decomposes at to form nitrogen and fluorine gases according to the following equation: 6.00-L reaction vessel is initially charged with 1.96 mol of NF₃ and allowed to come to equilibrium at 800 K.Once equilibrium is established,the reaction vessel is found to contain 0.0380 mol of N₂.What is the value of K_p at this temperature? (R = 0.0821 L.atm.mol.K)

Question

Quizplus · Accepted Answer

The equilibrium constant, $K_p$, can be calculated using the following expression:

$$K_p=\frac{(P_{N_2})(P_{F_2})^{3}}{(P_{NF_3})^{2}}$$

where $P_{N_2}$, $P_{F_2}$, and $P_{NF_3}$ are the partial pressures of each gas at equilibrium.

We are given the initial moles of NF$_3$ and the moles of NF$_3$ at equilibrium, so we can calculate the moles of N$_2$ and F$_2$ that were produced:

$\Delta$n(N$_2$) = 2(0.0380 mol) = 0.0760 mol

$\Delta$n(F$_2$) = 3(0.0380 mol) = 0.114 mol

Using the ideal gas law, we can then calculate the partial pressures of each gas:

$P_{N_2} = \frac{n_{N_2}}{V} 	imes RT$ = (0.0760 mol / 0.00600 L) x (0.0821 L atm/mol K) x (800 K) = 9.87 atm

$P_{F_2} = \frac{n_{F_2}}{V} 	imes RT$ = (0.114 mol / 0.00600 L) x (0.0821 L atm/mol K) x (800 K) = 14.8 atm

$P_{NF_3} = (1.96 mol - 0.0380 mol) / 0.00600 L 	imes 0.0821 L atm/mol K 	imes 800 K = 101 atm$

Substituting these values into the expression for $K_p$ gives:

$$K_p=\frac{(9.87)(14.8)^{3}}{(101)^{2}} = 1.91 	imes 10^{-3} 	ext{ atm}^{-2}$$

Converting from atm$^{-2}$ to mol$^{-2}$ using the ideal gas law gives:

$$K_c = K_p(RT)^{\Delta n} = 1.91 	imes 10^{-3} 	ext{ atm}^{-2} (0.0821 	ext{ L atm mol}^{-1} 	ext{K}^{-1} 	imes 800	ext{ K})^{2}$$

$$K_c = 1.76	imes 10^{3} 	ext{ mol}^{-2}$$

Therefore, the answer is (B) 1.91$	imes$10$^{-3}$atm$^{-2}$ or 1.76$	imes$10$^{3}$mol$^{-2}$.

Nitrogen Trifluoride Decomposes at to Form Nitrogen and Fluorine Gases ×\times×

Nitrogen Trifluoride Decomposes at to Form Nitrogen and Fluorine Gases $\times$