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Question 2
Nitrogen and hydrogen were mixed in a 1:3 mole ratio and left to reach equilibrium in a flask at a temperature of 550 K. The equation for the reaction between nitrog... show full transcript
Step 1
Answer
Given that the mole fraction of NH3 is 0.80, we can find the mole fractions of the other gases. Since the total pressure is 150 kPa, we can express the mole fraction of hydrogen as follows:
Let the mole fraction of H2 be denoted as x:
Mole fraction of nitrogen = 1 - 0.80 - x
Using the fact that the total mole fraction sums to 1, we have:
0.20 - x = mole fraction of nitrogen
Calculating the partial pressure of nitrogen:
Partial pressure of N2 = (mole fraction of N2) × (total pressure) Partial pressure of N2 = (0.20 - x) × 150 kPa
At equilibrium, using the stoichiometry of the reaction, we find: Partial pressure of NH3 = 0.80 × 150 kPa = 120 kPa Substituting: Partial pressure of H2 = P (total) - P (NH3) - P (N2) This can be solved accordingly.
Step 2
Answer
Partial pressure of H2 can be calculated using the relationship derived from mole fractions: Partial pressure of H2 = (mole fraction of H2) × (total pressure) Using a ratio from the stoichiometry: Partial pressure of hydrogen = 0.75 × 150 kPa = 112.5 kPa.
Step 3
Step 4
Step 5
Answer
Given the partial pressures from Table 2:
Kc = \frac{(1.10 × 10^2)^2}{(1.20 × 10^2) (1.50 × 10^2)^3}
Calculating gives Kc in units of (kPa)^{-2}.
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