When the following equilibrium process CO + 2H2(g) =CH2OH()
reached equilibrium in a particular experiment, the following equilibrium
concentrations were present.
[CO]=0.150M
[Hz]=0.360M
(CH3OH]=0.282M
Calculate the value of keq
?

K(eq) = 15  (2 sig. figs)

Explanation:

Rxn:         CO(g)    +    2H₂(g)    ⇄   CH₃OH(l)

C(eq):    0.150M       0.360M           0.282M  

Keq = [CH₃OH(l)]/[CO(g)][H₂(g)]

= (0.282M)/(0.150M)(0.360M)²  

= 14.50617284 (calc. ans.)

= 15 (2 sig. figs.)  

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.