Question

Be sure to answer all parts. Consider the following reaction at equilibrium: A(g) ⇆ 2B(g) From the data shown here, calculate the equilibrium constant (both KP and Kc) at each temperature. Is the reaction endothermic or exothermic? Temperature (°C) [A] (M) [B] (M) 200 0.0180 0.870 300 0.150 0.790 400 0.240 0.695 Kc(200°C) = KP(200°C) = Kc(300°C) = KP(300°C) = Kc(400°C) = KP(400°C) = Cannot be determined. The reaction is exothermic. The reaction is endothermic

Answer 1

Temperature(T c)    [A]       [B]

200               0.018      0.87

300             0.15     0.79

400            0.24     0.695

at 200 C ,   kC = [B]^2/[A]

    Kc = 0.87^2/0.018   = 42.05

KP = kC*(RT)^Dn

Dn = 2-1= 1

Kp = 42.05*(0.0821*473.15)^1

   = 1633.46

at 300 C

kC = [B]^2/[A]

    Kc = 0.79^2/0.15   = 4.16

KP = kC*(RT)^Dn

Dn = 2-1= 1

Kp = 4.16*(0.0821*573.15)^1

   = 195.75

at 400 c

kC = [B]^2/[A]

    Kc = 0.695^2/0.24   = 2.013

KP = kC*(RT)^Dn

Dn = 2-1= 1

Kp = 2.013*(0.0821*673.15)^1

      = 111.25

by ln(k2/k1) = DH/R [ 1/T1-1/T2]

(ln(2.013/(4.16))= x/(8.314)(1/573-1/673),

DHrxn = -23.272 kj/mol

it is exothermic