Calculate the equilibrium constant (Kc) for the reaction at this temperature. If

Question

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

If you could be explicit in how you came to your answer it would be greatly appreciated. Thank you

Part A Calculate the equilibrium constant (Kc) for the reaction at this temperature Consider the following reaction: CO(g) 2 H2(9) CH3 OH(g) A reaction mixture in a 5.22 -L flask at a certain temperature initially contains 27.0 g CO and 2.32 g H2. At equilibrium, the flask contains 8.64 g CH3 OH Kc Submit My Answers Give Up

Explanation / Answer

According to the Equilibrium constant

For the calululation Lower concentrated reactants and products will be considered (all are in gaseous state that means lower concentrated).

Kc = [CH3OH]/[CO][H2]2

Concentration of CO: w =27g (according to the question), molecular weight = 28, V = 5.22L

M = wt/mwt x1/v in litre

M = 27/28 x1/5.22 = 0.1847 M

H2 Concentration: w=2.32 g, molecular weight = 2, V =5.22L

M= 2.32/2 x 1/5.22

=0.222M.

CH3OH concentration: w=8.64 g, molecular weight= 32 V =5.22L

M= 8.64/32 x 1/5.22

= 0.0517M.

The Equilibrium constant (Kc) = [CH3OH]/[CO][H2]2

= 0.0517/0.1847 x [0.222]2

=0.0517/0.1847 x 0.0492

= 0.2799 / 0.0492

= 5.689

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