For the following reaction, K c = 0.513 at 500 K . N 2 O 4 ( g ) ? 2 N O 2 ( g )

Question

For the following reaction, Kc = 0.513 at 500 K.
N 2O 4(g)?2NO 2(g) Part A If a reaction vessel initially contains an N 2O 4 concentration of 5.50 For the following reaction, Kc = 0.513 at 500 K.
N 2O 4(g)?2NO 2(g) For the following reaction, Kc = 0.513 at 500 K.
N 2O 4(g)?2NO 2(g) For the following reaction, Kc = 0.513 at 500 K.
N 2O 4(g)?2NO 2(g) For the following reaction, Kc = 0.513 at 500 K.
N 2O 4(g)?2NO 2(g) Part A If a reaction vessel initially contains an N 2O 4 concentration of 5.50 Part A If a reaction vessel initially contains an N 2O 4 concentration of 5.50 Part A If a reaction vessel initially contains an N 2O 4 concentration of 5.50 Part A If a reaction vessel initially contains an N 2O 4 concentration of 5.50 Part A If a reaction vessel initially contains an N 2O 4 concentration of 5.50 Part A If a reaction vessel initially contains an N 2O 4 concentration of 5.50 For the following reaction, Kc = 0.513 at 500 K.
N 2O 4(g)?2NO 2(g) Part A If a reaction vessel initially contains an N 2O 4 concentration of 5.50

Explanation / Answer

N2O4(g) ?2NO2(g)

5.5*10^-2 - x 2x At equillibrium.

Kc= (2x)^2/(5.5*10^-2 - x)

0.513 = (2x)^2/(5.5*10^-2 - x)

Solving thiis

x=0.0415

So equilbrium conc. of NO2 = 2x = 0.083 M

equilbrium conc. of N2O4 = 5.5*10^-2 - x = 0.0135 M.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.