Constants | Periodic Table Part A The N204, NO2 equilibrium mixture in the flask

Question

Constants | Periodic Table Part A The N204, NO2 equilibrium mixture in the flask on the left in the figure is allowed to expand into the evacuated flask on the right. N204 (g) 2NO2 (g): Kc 4.61 x 10-3 at 25 °C What is the composition of the gaseous mixture when equilibrium is re-established in the system consisting of the two flasks? Calculate the amount (in moles) of NO2. Express your answer with the appropriate units (Figure 1) Hyo,=| Value ! Units Submit Request Answer Part B Figure 1of1 Calculate the amount (in moles) of N204 Express your answer with the appropriate units VN204- 971 mol N,O 0.0580 mol NO2 0.750 L 25°C 2.25 L 25 Submit Request Answer

Explanation / Answer

as the volume of gas is increasing, the partial pressures decreases.

so that, equilibrium shift towards more No of moloes side.

           N2O4(g) <===> 2NO2(g)

initial 0.971/0.75       0.058/0.75

          = 1.295 M       = 0.0773 M

after changing volume

   1.295*0.75/(2.25+0.75)   0.0773*0.75/(2.25+0.75)

              0.324 M        0.0193 M

chnage         - x             +2x

equil      0.324-x            0.0193+2x

Kc = [NO2]^2/[N2O4]

(4.61*10^-3) = (0.0193+2x)^2/(0.324-x)

X = 0.0094

part A : nO of mol of NO2 = (0.0193+2*0.0094)*3 = 0.1143 mol

part B : nO of mol of N2O4 = (0.324-0.0094)*3 = 0.9438 mol

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