For the reaction 2 NO₂(g)⇌ N₂O₄(g)K_p equals ________.

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is K = [NO₂]²/[N₂O₄]. Taking the natural logarithm of both sides, we get ln K = 2 ln[NO₂] - ln[N₂O₄]. This can be rearranged to give ln K = -ln[N₂O₄] + 2 ln[NO₂]. Applying the Arrhenius equation, ln K = -Ea/RT - ln[A] + ln[B], where Ea is the activation energy, A and B are the pre-exponential factors, and R is the gas constant. Since ln K = -ln[N₂O₄] + 2 ln[NO₂], we can identify ln[N₂O₄] with -Ea/RT - ln[A], and ln[NO₂] with ln[B]. Therefore, ln[N₂O₄]/RT = Ea/RT² + ln[A]. Taking the exponential of both sides, we get N₂O₄ = A exp(-Ea/RT) / RT. Substituting this expression into the equilibrium constant expression, we get K = [NO₂]² A exp(-Ea/RT) / [RT]³[N₂O₄]. Rearranging terms, we get ln K = -ln[R] - ln[T] - 3 ln[N₂O₄] + 2 ln[NO₂] + ln[A] -Ea/RT. Therefore, ln K = ln(K_c/RT) = -Ea/RT + ln(K_c), which gives us the desired result: K_p = exp(-Ea/RT) / R.

For the Reaction 2 NO2(g)⇌ N2O4(g)Kp Equals ________

For the Reaction 2 NO₂(g)⇌ N₂O₄(g)K_p Equals ________