Answer to Question #128681 in General Chemistry for A

Question

Good day,

where R=8.314 L⋅kPa/(K⋅mol)
For the reaction
3A(g)+2B(g)⇌C(g)
Kc= 72.2 at a temperature of 237 ∘C
Calculate the value of Kp
Express your answer numerically.I calculated 2.34x10^-5 but says incorrect,please help

Thanks

Answer 1

K_p = K_c×(R×T)^Δn

Be careful about the value of R. To decide which value of R to use when you convert between Kc and Kp, you can reason as follows. Kc involves molar concentrations, for which the units are mol/L; Kp involves pressures expressed in atm or Pa.

In this case we will use R=8.314 L⋅kPa/(K⋅mol)

There is 1 mol of gas on the product side of the chemical equation, and there are 3+2=5 mol of gaseous reactants, so Δn = (1 mol - 5 mol) = -4 in this case.

T = 237 + 273.15 = 510.15 K

K_c= 72.2

K_p = 72.2×(8.314×510.15)^-4 = 72.2×(8.314×510.15)^-4 = 72.2×(4241.38)^-4 = 72.2×3.09 ×10^-15 = 2.23×10^-13 kPa^-4

Answer: 2.23×10^-13 kPa^-4