Consider the following reversible reaction.

Upper C upper O (g) plus 2 upper H subscript 2 (g) double-headed arrow upper C upper H subscript 3 upper O upper H (g).

What is the equilibrium constant expression for the given system?
K e q equals StartFraction StartBracket upper C upper O EndBracket StartBracket upper H 2 EndBracket superscript 2 over StartBracket upper C upper H subscript 3 upper O upper H EndBracket EndFraction.
K e q equals StartFraction StartBracket upper C upper H subscript 3 upper H upper O EndBracket over StartBracket upper C upper O EndBracket StartBracket upper H 2 EndBracket superscript 2 EndFraction.
K e q equals StartFraction StartBracket upper C upper O EndBracket StartBracket upper H 2 EndBracket over StartBracket upper C upper H subscript 3 upper O upper H EndBracket EndFraction.
K e q equals sart fraction StartBracket upper C upper H subscript 3 upper O StartBracket upper C upper O EndBracket StartBracket upper H 2 EndBracket EndFraction.

metric time was an attempt to create a standard unit of time based off of 10, but it never took off an is not included in the metric system (si)

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole