For chemical reactions involving ideal gases, the equilibrium constant K can be

Question

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.'

Part A

For the reaction

2CH4(g)C2H2(g)+3H2(g)

K = 0.130 at 1668 C . What is Kp for the reaction at this temperature?

Express your answer numerically

Part B

For the reaction

N2(g)+3H2(g)2NH3(g)

Kp = 3.60×103 at 272 C . What is K for the reaction at this temperature?

Enter your answer numerically.

Explanation / Answer

M = mol per liter

P = pressure

we can always relate concnetration and pressure with ideal gas law

PV = nRT

n = mass/MW

PV = mRT/MW

M = m/V = P*MW/(RT)

Then

Kp = Kc*(RT)^(nf-ni)

R = gas constnatn, must be in mol per liter i.e. 0.082

nf = final amount of moles in productsd

ni = initial amount of moles in reactants

therefore

for A

nf = 3mol ofH2 + 1 mol of C2H2

ni = 2 mol of CH4

nf-ni = 4-2 = 2

then

Kp = Kc*(RT)^2

NOTE: T is not 298 but (1668 + 273) = 1941

Kp =(0.130)(0.082*1941)^2 = 3293.230

B)

nf = 2

ni = 4

nf.ni = 2-4 = -2

then

Kp = Kc*(RT)^-2

Kc = Kp*(RT^2)

NOTE T is no 298 but 272 + 273 = 545

Kc = (3.6*!0^-3)(0.082*545)^2 = 7.1899

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