At 700 K, the equilibrium constant Kp, for the reaction 2SO3(g)⇌2SO2(g)+O2(g) is 1.8×10−3 kPa. What is the numerical value on moles per dm3 of Kc for the reaction at the same temperature ?
At 700 K, the equilibrium constant Kp, for the reaction 2SO3(g)⇌2SO2(g)+O2(g) is 1.8×10−3 kPa. What is the numerical value on moles per dm3 of Kc for the reaction at the same temperature ?
The correct option is B 3.09×10−7
Given reaction scheme
2SO3(g)⇌2SO2(g)+O2(g)
∆ng=
nproducts-nreactants =3−(2+1)=3−2=1
Formula: Kp=Kc(RT)Δn
Kp
is equilibrium constant from partial pressures = 1.8 kpa = 1.8×103pa
Kc is the equilibrium constant from concentrations =?
R is gas constant = 8314 L pa K-1 mol-1
T is
temperature = 700K
On substituting we get
Kc=1.8×103pa8314LpaK−1mol−1×700K
Kc=3.09×10−7molL−1
Hence
option B is correct.
Given reaction scheme
2SO3(g)⇌2SO2(g)+O2(g)
∆ng= nproducts-nreactants =3−(2+1)=3−2=1
Formula: Kp=Kc(RT)Δn
Kp is equilibrium constant from partial pressures = 1.8 kpa = 1.8×103pa
Kc is the equilibrium constant from concentrations =?
R is gas constant = 8314 L pa K-1 mol-1
T is temperature = 700K
On substituting we get
Kc=1.8×103pa8314LpaK−1mol−1×700K
Kc=3.09×10−7molL−1
Hence option B is correct.
