The equilibrium constant, Kc, for the reaction: SO3 + NO <--> NO2 + SO2?
was found to be 0.500 at a certain temperature. if 0.300 mol of SO3 and 0.300 mol of NO were placed in a 1.00 L container and allowed to react, what would be the equilibrium concentration of each gas?
Answer:
Initial concentration SO3 = 0.300 mol/ 1.00 L = 0.300 M
Initial concentrattion NO = 0.300 mol/ 1.00 L = 0.300 M
at equilibrium
[ SO3 ]= 0.300 - x
[ NO ] = 0.300 - x
[ NO2 ] = x
[ SO2 ] = x
K = [ NO2 ] [ SO2 ] / [ SO3 ] [ NO ]
K = 0.500 = (x)(x( / (0.300 - x)( 0.300 - x)
We can solve this equation by squaring both sides
sqr 0.500 =0.707
0.707 = x / 0.300 - x
x = 0.124 M
[SO3 ] = [NO ] =0.300 - 0.124 = 0.176 M
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Answer:
Initial concentration SO3 = 0.300 mol/ 1.00 L = 0.300 M
Initial concentrattion NO = 0.300 mol/ 1.00 L = 0.300 M
at equilibrium
[ SO3 ]= 0.300 - x
[ NO ] = 0.300 - x
[ NO2 ] = x
[ SO2 ] = x
K = [ NO2 ] [ SO2 ] / [ SO3 ] [ NO ]
K = 0.500 = (x)(x( / (0.300 - x)( 0.300 - x)
We can solve this equation by squaring both sides
sqr 0.500 =0.707
0.707 = x / 0.300 - x
x = 0.124 M
[SO3 ] = [NO ] =0.300 - 0.124 = 0.176 M
[ NO2 ] = [ SO2 ] = 0.124 M