Question

A United Nations toxicologist studying the properties of mustard gas, S(CH2CH2CI)2, a blistering agent used in warfare, prepares a mixture of 0.675 M SCI, and 0.973 M C H, and allows it to react at room temperature (20.0°C): SC12(g) + 2 C2H4() = S(CH,CH,CI), () At equilibrium,[S( CH,CH,CI)21 -0.350 M. Calculate Kp.

Answer 1

making ICE table- S(CH₂ CH₂ (l), (g) 1 SC2(g) + 2 C₂ Hu (g) = 0.675M 0.973m - a © -22 tu 0.675- 0.973-22 a kc = [S(CH₂ CH₂ CL 22 ) [ 52] [C2 Hu]² Acc. to Ques, at equilibrium- [S(CH ₂ CH ₂ (0)2] = x=0.350m . At Equilibrium - [su2] = 0.675 - x = 0.675 -0.50 = 0.325 m (C2H4 ) = 0.973 - 2x = 0.973-210.350) = 0.273 m plugging values in equ l to calculate ke as- Ke & (0.350) - 14.45 (0.325) (0.273)2 > no. of gaseous moles in reactants Also, (ne - nr) Kp = kc (RT) AY no. of gascous moles in product Temperature Gas Constant Here we have - T= 20°c = (273 +20k = 293k An= {1]-[1+2] = -2 R = 0.0821 Lath mollk" Kc = 14.45

plugging values, we get - kp= 14.45 (0.0821 x 29 30 = Kp = 14.45 (24.06)2 0.025 =) kp = Please Rate...