Relating Different Forms of the Equilibrium Constant For chemical reactions invo
ID: 837743 • Letter: R
Question
Relating Different Forms of the Equilibrium Constant For chemical reactions involving ideal gases, the equilibrium constant K can be expressed either in terms of the concentrations of the gases (in M) or as a function of the partial pressures of the gases (in atmospheres). In the latter case, the equilibrium constant is denoted as Kp to distinguish it from the concentration-based equilibrium constant K. For the reaction K = 0.135 at 1572 degree C. What is Kp for the reaction at this temperature? Express your answer numerically. For the reaction Kp = 5.00 Times 10-3 at 342 degree C. What is K for the reaction at this temperature? Enter your answer numerically.Explanation / Answer
(A) 2 CH4 <=> C2H2 + 3 H2
Change in moles of gas Dn = 1 + 3 - 2 = 2
T = 1572 deg C = 1845.15 K
R = 0.08206 atm.L/mol.K
Kp = K x (RT)^(Dn)
= 0.135 x (0.08206 x 1845.15)^2
= 3095 = 3.09 x 10^3
(B) N2 + 3 H2 <=> 2 NH3
Change in moles of gas Dn = 2 - 1 - 3 = -2
T = 342 deg C = 615.15 K
R = 0.08206 atm.L/mol.K
K = Kp x (RT)^(-Dn)
= 5.00 x 10^(-3) x (0.08206 x 615.15)^(2)
= 12.74 = 12.7