Question #99506

A 4.00 L flask was filled with 12.00 mol of SO2 and 2.00 mol of NO2. After equilibrium was reached, it was found that 11.30 mol of NO was present. Calculate the value of all the equilibrium concentrations and calculate the value of Kc.

Expert's answer

SO2+NO2NO+SO3SO_2 + NO_2 \leftrightarrow NO + SO3

c=nVc=\frac{n}{V}


cinitial(SO2)=12.004.00=3.00molLc_{initial} (SO2) = \frac{12.00}{4.00} = 3.00 \frac{mol}{L}


cinitial(NO2)=12.004.00=3.0molLc_{initial}(NO_2) = \frac{12.00}{4.00} = 3.0 \frac{mol}{L}


ceq(NO)=11.304.00=2.825molLc_{eq}(NO) = \frac{11.30}{4.00}= 2.825\frac{mol}{L}


SO2SO_2 + NO2NO_2 \leftrightarrow NONO + SO3SO_3

I: 3.00 3.00 0 0

C: -2.825 -2.825 +2.825 +2.825

E: 0.175 0.175 2.825 2.825



Kc=[NO][SO3][SO2][NO2]=2.825×2.8250.175×0.175=261K_c = \frac{[NO][SO_3]}{[SO_2][NO_2]} = \frac{2.825\times2.825}{0.175\times0.175} = 261


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