Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium pressure of D2 (deuterium, an isotope of hydrogen).

D2(g) + H2(g) ⇌ 2 HD(g) Kp = 1.80

P(H2)eq = 1.1 × 10-3 atm P(HD)eq = 2.7 × 10-3 atm

Question

Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium pressure of D2 (deuterium, an isotope of hydrogen).

D2(g) + H2(g) ⇌ 2 HD(g)        Kp = 1.80

P(H2)eq = 1.1 × 10-3 atm          P(HD)eq = 2.7 × 10-3 atm

BlackTom AI · Accepted Answer

For the equilibrium D2(g) + H2(g) ⇌ 2 HD(g), the equilibrium-constant expression in terms of pressure (Kp) is based on partial pressures and stoichiometry.

1. Relevant knowledge points and concepts
- For a reaction aA + bB ⇌ cC, Kp = (P_C^c)/(P_A^a P_B^b).
- Here stoichiometry gives Kp = (P_HD)^2 / (P_H2 · P_D2).
- To find an unknown partial pressure, rearrange the Kp expression algebraically.
- Keep track of units (atm) and scientific notation when squaring small numbers.

2. Step-by-step reasoning
Start with the Kp expression for this reaction:
Kp = (P_HD)^2 / (P_H2 · P_D2).

Solve for P_D2:
P_D2 = (P_HD)^2 / (Kp · P_H2).

3. Calculations with intermediate steps
Given: P_HD = 2.7 × 10^-3 atm, P_H2 = 1.1 × 10^-3 atm, Kp = 1.80.
Square P_HD:
(P_HD)^2 = (2.7 × 10^-3)^2 = 2.7^2 × (10^-3)^2 = 7.29 × 10^-6 atm^2.
Compute denominator:
Kp · P_H2 = 1.80 × 1.1 × 10^-3 = 1.98 × 10^-3 atm.
Divide:
P_D2 = (7.29 × 10^-6) / (1.98 × 10^-3) = (7.29/1.98) × 10^-3 ≈ 3.6818 × 10^-3 atm.
Round to two significant figures (consistent with given data): 3.7 × 10^-3 atm.

4. Why this answer is correct
The algebraic rearrangement uses the correct Kp expression and accounts for stoichiometric coefficients (the HD term is squared). Numerical substitution and careful handling of powers of ten yield the computed P_D2.

5. Why the other options are incorrect
- 2.7 atm, 1.4 atm, 0.73 atm: off by several orders of magnitude; they ignore the small equilibrium pressures given (10^-3 atm).
- 8.1 × 10^-4 atm: too small; it would result from a different algebraic manipulation or arithmetic error (not from the correct Kp rearrangement).

6. Conclusion
Calculated equilibrium pressure P(D2) = 3.7 × 10^-3 atm, which matches the provided correct answer.

Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium pressure of D2 (deuterium, an isotope of hydrogen). D2(g) + H2(g) ⇌ 2 HD(g) Kp = 1.80 P(H2)eq = 1.1 × 10-3 atm P(HD)eq = 2.7 × 10-3 atm单项选择题

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