Q3:a) the reaction: N204(g) 5 2NO2g) Kc = 4.66 x 104 mol/L At temperature - 80.0 °C the total pressure of both gases was found to be 0.79 atm. Calculate the pressure of each of the two gases at equilibrium. Detailed calculations {Tips: Assume PNo2 = X atm . so at equilibrium, the N,0, = total - 2X. To get out of the quadratic equation and solve for X, you can ignore the 2X since it is too small relative to the total. First, you need to convert Kc to Kp in order to calculate the pressure; P}

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Q3:a) the reaction: N204(g) 5 2N02(g) Kc = 4.66 x 104 mol/L
At temperature - 80.0 °C the total pressure of both gases was found to be 0.79 atm.
Calculate the pressure of each of the two gases at equilibrium. Detailed calculations
{Tips: Assume PNo2 = X atm . so at equilibrium, the N,0, = total - 2X. To get out of the quadratic equation and solve for X,
you can ignore the 2X since it is too small relative to the total. First, you need to convert Kc to Kp in order to calculate
the pressure; P}
Transcribed Image Text:Q3:a) the reaction: N204(g) 5 2N02(g) Kc = 4.66 x 104 mol/L At temperature - 80.0 °C the total pressure of both gases was found to be 0.79 atm. Calculate the pressure of each of the two gases at equilibrium. Detailed calculations {Tips: Assume PNo2 = X atm . so at equilibrium, the N,0, = total - 2X. To get out of the quadratic equation and solve for X, you can ignore the 2X since it is too small relative to the total. First, you need to convert Kc to Kp in order to calculate the pressure; P}
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