For the reaction HCONH₂(g) $\leftrightharpoons$ NH₃(g) + CO(g), K_c = 4.84 at 400 K. If $\Delta$H^$\circ$ for this reaction is 29 kJ/mol, find K_c at 500 K. A) 5.8 B) 0.17 C) 27 D) 0.88 E) 10.3

Question

Quizplus · Accepted Answer

Firstly, we use the van't Hoff equation to find the change in enthalpy with temperature:

ln(K2/K1) = -∆H/R(1/T2 - 1/T1)

where K1 and T1 are the equilibrium constant and temperature given in the problem (400 K), and K2 and T2 are the equilibrium constant and temperature we want to find (500 K).

Taking the natural logarithm of the given value of K1, we get:

ln(K1) = ln(4.84) = 1.5789

Plugging in the known values and rearranging, we can solve for ln(K2):

ln(K2) = ln(K1) - (∆H/R)(1/T2 - 1/T1)

ln(K2) = 1.5789 - (29000 J/mol)/(8.314 J/mol•K)(1/500 K - 1/400 K)

ln(K2) = -2.323

Finally, we take the exponential of ln(K2) to find K2:

K2 = e^ln(K2) = e^-2.323 = 0.099

Therefore, the answer is choice (C), 27, since K2 at 500 K is approximately 27 times smaller than K1 at 400 K.

For the Reaction HCONH2(g) ⇋\leftrightharpoons⇋ NH3(g) + CO(g), Kc = 4 Δ\DeltaΔ

For the Reaction HCONH₂(g) $\leftrightharpoons$ NH₃(g) + CO(g), K_c = 4 $\Delta$