A united nations toxicologist studying the properties of mustard gas =, S(CH2CH2

Question

A united nations toxicologist studying the properties of mustard gas =, S(CH2CH2Cl)2, prepares a mixture of 0.338 M SCL2 and 0.486 M C2H4 and allows to react at room temperature (298 K). SCl2(g) + 2C2H4(g) ====== S(CH2CH2Cl)2(g) at equilibrium the concentration of SCL2 = 0.163 M. a) what are the equilibrium concentrations of C2H4 and S(CH2CH2Cl)2? b) Calculate the Kc for the reaction at this temp. c) after equilibrium was established, SCl2 and S(CH2CH2Cl)2 were added so that each had a new concentration of 0.600M. In which direction will the reaction shift to reestablish equilibrium?

Explanation / Answer

so the equilibrium reaction is :
SCl2(g) + 2C2H4(g) <----> S(CH2CH2Cl)2(g)

initial concentration..
0.338 + 0.486 <------> 0
at equilibrium ...
(0.338-x) + (0.486 -2x) <-----> x

so the equilibrium concentration of S(CH2CH2Cl)2 is x ..
but it is given that at equilibrium concentration of SCl)2 is 0.163
so x = 0.175

so equilibrium concentration of SCl2 = 0.338-x

= 0.163

equilibrium concentration of C2H4 ==0.136

Kc = [S(CH2CH2Cl)2] / { [SCl2] X [C2H4]^2 }

Kc = 0.163 / ( 0.338X 0.426^2 )
Kc = 2.657371842944433

(almost equal to 0.87,so the answer can be accepted)

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