Chemistry, 03.03.2020 03:22, icantspeakengles
Some CH2Cl2 is placed in a sealed flask and heated to 517 K. When equilibrium is reached, the flask is found to contain CH2Cl2 (3.42×10-2 M), CH4 (3.69×10-2 M), and CCl4 (4.12×10-2 M). What is the value of the equilibrium constant for this reaction at 517 K?
Answers: 2
follows:
for above equation follows:![K_{eq}=\frac{[CH_4][CCl_4]}{[CH_2Cl_2]^2}](/tpl/images/0531/6381/cdd40.png)
![[CH_4]_{eq}=3.69\times 10^{-2}M](/tpl/images/0531/6381/50f3b.png)
![[CCl_4]_{eq}=4.12\times 10^{-2}M](/tpl/images/0531/6381/3e13d.png)
![[CH_2Cl_2]_{eq}=3.42\times 10^{-2}M](/tpl/images/0531/6381/c3f41.png)
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Some CH2Cl2 is placed in a sealed flask and heated to 517 K. When equilibrium is reached, the flask...
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