The equilibrium constant, Kc, is calculated using molar concentrations. For gase

Question

The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp = Kc(RT)delta n where R = 0.08206 L . atm/(K . mol), T is the absolute temperature, and An is the change in the number of moles of gas. For example, consider the reaction N2(g) + 3H2(g) double arrow 2NH3(g) for which An = 2 - (1 + 3) = -2. For the reaction 2A(g) + 2B(g)double arrow C(g) Kc = 91.4 at a temperature of 147 degree C. Calculate the value of Kp For the reaction X(g)+3Y(g) doublearrow 3Z(g) Kp= 1.27x10-2 at a temperature of215 degree C. Calculate the value of Kc.

Explanation / Answer

part a) Kp=Kc(RT)^ n

? n=1-(2+2)=-3 n=1-(2+2)=-3

Kp=Kc(RT)^? n Kp=Kc(RT)^? n Kp=Kc(RT)^ n

=94.1(0.08206*420)^-3 =94.1(0.08206*420)^-3 =94.1(0.08206*420)^-3

=2.298*10^-3 =2.298*10^-3 =2.298*10^-3

partb) partb) partb)

? n=3-(1+3) ? n=3-(1+3) ? n=3-(1+3) n=3-(1+3)

=-1 =-1 =-1 =-1

Kp=Kc(RT)^? n Kp=Kc(RT)^? n Kp=Kc(RT)^? n Kp=Kc(RT)^? n Kp=Kc(RT)^ n

Kc=Kp/ Kc=Kp/ Kc=Kp/ Kc=Kp/ Kc=Kp/ (RT)^? n

Kc=1.27*10^-2/(0.08206*488)^-1

=.508

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