1. The equilibrium constant, Kc, for the following
reaction is 42.2 at 288 K.
2CH2Cl2(g)
<-----> CH4(g) +
CCl4(g)
When a sufficiently large sample of
CH2Cl2(g) is introduced into
an evacuated vessel at 288 K, the equilibrium
concentration of CCl4(g) is found to be
0.140 M.
Calculate the concentration of
CH2Cl2 in the equilibrium
mixture. _____ M
2. A student ran the following reaction in the laboratory at
295 K:
2CH2Cl2(g)
<-------> CH4(g) +
CCl4(g)
When she introduced 8.46×10-2 moles of
CH2Cl2(g) into a 1.00 liter
container, she found the equilibrium concentration of
CCl4(g) to be
3.90×10-2M.
Calculate the equilibrium constant, Kc, she obtained for
this reaction.
Kc =
1. At equilibrium , [CH4 ] = [CCl4 ] = 0.140 M
and Kc = 42.2
For given reaction
Kc = [CH4][CCl4] / [CH2Cl2]2
42.2 . = (0.140)2 / [CH2Cl2]2
[CH2Cl2] = 0.0216 M
2. At equilibrium , [CH4 ] = [CCl4 ] = 3.90 *10-2 M = 0.039 M
Initial Moles of CH2Cl2 = 8.46 *10-2 moles = 0.0846 moles
Volume = 1L
so, initial [CH2Cl2 ] = moles / volume = 0.0846 moles/ 1 L = 0.0846 M
[CH2Cl2] reacted = 2 * [CCl4] = 2* 0.039 = 0.078 M
At equilibrium , [CH2Cl2] = Initial value - decomposed value = 0.0846 - 0.078 = 0.0066 M
Now, Kc = [CH4][CCl4] / [CH2Cl2]2
Kc = (0.039)2 / (0.0066)2 = 34.92
1. The equilibrium constant, Kc, for the following reaction is 42.2 at 288 K. 2CH2Cl2(g) <----->...
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