Question #4. Nitrogen gas and hydrogen gas are combined and allowed to react at

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Question #4. Nitrogen gas and hydrogen gas are combined and allowed to react at a certain temperature according to the following reaction: N2(g) + 3H2(g) double arrow 2 NH3(g)

At equilibrium the total pressure in the vessel is 50.0 atm. The mixture contains the following mole fractions: 0.890 NH3, 0.228 N2, and 0.683 H2. Determine Kp at the temperature of this experiment.

Nitrogen gas and hydrogen gas are combined and allowed to react at a certain temperature according to the following reaction: N2(g) + 3 H2(g) 2 NH3 (g) At equilibrium the total pressure in the vessel is 50.0 atm. The mixture contains the following mole fracions: 0.890 NHs, 0.228 N2, and 0.683 H2. Determine Kp at the temperature of this experiment. N 2 5. Consider the following reaction: 2.50 moles/L of N204 was placed into a sealed vessel and allowed to come to equilibrium at a constant temperature. At equilibrium, 72.3% of the N204(g) had been converted to NO(g). Calculate Ke at the temperature of this experiment. Consider the fallowing reaction and equilibrium constant at a given temperature: NH.cds) NH3(g) + HCI(g) K.-3.5 x 10 A 2.00 L reaction vessel contains 0.0200 moles of NH3(9),0.050 moles HCI(g), and 0.120 moles of NH.CI(s). In which direction will reaction proceed to attain equilibrium? Supporting calculations are required. 2L 2.059 mdl 0.0 25 M 0.050 ml0.0 25 M

Explanation / Answer

You have given incorrect data of mole fractions, because they do not add to give 1. Kindly verify.

Total Pressure, PT = 50 atm

First calculate partial pressures:

Partial pressure of H2, pH2 = XH2*PT

Partial pressure of N2, pN2 = XN2*PT

Partial pressure of NH3, pNH3 = XNH3*PT

For this reaction, we have the following equation for Kp:

Kp = (pNH3)2/( (pH2)3*(pN2) )

Put values and solve.

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