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A toxicologist studying mustard gas, S(CH2CH2Cl)2, a
Chapter 17, Problem 44P(choose chapter or problem)
Problem 44P
A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SC12 and 0.973M C2H4 and allows it to react at room temperature (20.0°C):
SCl2 (g) + 2C2H4( g) ⇌ S (CH2CH2Cl) 2 (g)
At equilibrium, [S(CH2CH2C1) 2] = 0.350M. Calculate Kp.
Questions & Answers
QUESTION:
Problem 44P
A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SC12 and 0.973M C2H4 and allows it to react at room temperature (20.0°C):
SCl2 (g) + 2C2H4( g) ⇌ S (CH2CH2Cl) 2 (g)
At equilibrium, [S(CH2CH2C1) 2] = 0.350M. Calculate Kp.
ANSWER:
Solution 44P:
Here, we are going to calculate the Kp for the reaction.
Step 1:
Given reaction
SCl2 (g) + 2C2H4( g) ⇌ S (CH2CH2Cl)2 (g)
Given that,
[SCl2] =0.675 M
[C2H4] =0.973 M
At equilibrium, [S(CH2CH2Cl)= 0.350M
Hence, the reaction table for the reaction is given below:
Therefore,
The concentrations at equilibrium are
[S(CH2CH2Cl)] =
[S(CH2CH2Cl)] =0.350M
[SCl2] = (0.675-)
[SCl2] = (0.675-0.350) = 0.325 M
[C2H4] =(0.973-x)
[C2H4] =(0.973-20.350) M=0.273 M