Question

Enter your answer in the provided box. A United Nations toxicologist studying the properties of mustard gas, S(CH2CH2Cl)2, a blistering agent used in warfare, prepares a mixture of 0.675 M SCl2and 0.973 M C2H4and allows it to react at room temperature (20.0°C): SCl2(g) + 2 C2H4(g) ⇌ S(CH2CH2Cl)2(g) At equilibrium,[S( CH2CH2Cl)2] = 0.350 M. Calculate _______________Kp.

Answer 1

Answer:

$0.024$

Explanation:

First we have to use the ICE table to find the equilibrium concentrations:

$SCl_2_(_g_) + 2 C_2H_4_(_g_) S(CH_2CH_2Cl)_2_(_g_)$

I    0.675 M   0.973 M           0

C    -X               -2X               +X

E  0.675-X     0.973-2X          X

The equilibrium concentration is given by the problem:

X=0.35 M

Then we have to do the substractions:

$SCl_2_(_g_)$ =  0.325 M

$C_2H_4_(_g_)$ = 0.273 M

$S(CH_2CH_2Cl)_2_(_g_)$= 0.35 M

With the equilibrium concentrations is posible to calculate Kc:

$Kc=\frac{[S(CH_2CH_2Cl)_2]}{[SCl_2][C_2H_4]^2}=\frac{[0.35]}{[0.325][0.273]^2}=14.45$

With the Kc value is posible calculate Kp using the equation:

Kp = Kc(RT)^(Δn)

Kp= Equilibrium Pressure constant

Kc= Equilibrium Concentration constant

R= 0.08206 L*atm/mol*K

T= Temperature

Δn=moles of products - moles of reactants

T= 25 ºC= 298.15 K

Δn = 1 mole of products $S(CH_2CH_2Cl)_2_(_g_)$- 3 moles of reactants $2 C_2H_4_(_g_)~and~SCl_2_(_g_)$= -2

$Kp=~14.45~(0.082*298.15)^(^-^2^)= 0.024$

Answer 2

Answer:

Kp = 2.04

Explanation:

SCl₂(g) + 2 C₂H₄(g) ⇌ S(CH₂CH₂Cl)₂(g)

Kp = [S(CH₂CH₂Cl)₂] / [SCl₂][C₂H₄]²

[S(CH₂CH₂Cl)₂] = 0.350 M

[SCl₂] = [S(CH₂CH₂Cl)₂] = 0.350 M

[C₂H₄] = 2 x [S(CH₂CH₂Cl)₂] = 2 x 0.350 M = 0.700 M

Kp = 0.350 / 0.350 x 0.700²

Kp = 2.04