7.2 What is K for the following equ...
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7.2 What is K for the following equilibrium when the equilibrium concentration of each substance is: (SO )= 0.60M, (0.) = 0.82M and (SO) = 1.90M ? 280,(g) + 0,9 = 250,(g) At a certain temperature and total pre 7.3

NEET/Medical Exams
Chemistry
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The equilibrium constant ( left(mathrm{K}_{mathrm{C}}right) ) for the give reaction is: [ begin{aligned} K_{c} &=frac{left[mathrm{SO}_{3}right]^{2}}{left[mathrm{SO}_{2}right]^{2}left[mathrm{O}_{2}right]} &=frac{(1.90)^{2} mathrm{M}^{2}}{(0.60)^{2}(0.821) mathrm{M}^{3}} end{aligned} ] ( =12.239 mathrm{M}^{-1}(text {approximately }) ) Hence, ( K ) for the equilibrium is ( 12.239 mathrm{M}^{-1} )
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