More practice questions

  1. Q4. NCERT’s experiment 3 at 731 K started with 2.44 × 10−2 M H2 and 1.98 × 10−2 M I2 and reached [HI] = 3.34 × 10−2 M. Find [H2] and [I2] at equilibrium, then Kc.

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    Each used ½ × 3.34 = 1.67, so [H2] = 2.44 − 1.67 = 0.77 and [I2] = 1.98 − 1.67 = 0.31 (all × 10−2 M). Kc = (3.34 × 10−2)2/{(0.77 × 10−2)(0.31 × 10−2)} = (1.116 × 10−3)/(2.387 × 10−5), about 46.7.

  2. Q5. Write Kc for 2NH3(g) + 5/2 O2(g) ⇌ 2NO(g) + 3H2O(g), and relate it to Kc for 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g).

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    Kc = [NO]2[H2O]3/[NH3]2[O2]5/2. The equation is half of NCERT’s, so this constant is the square root of the constant for the full equation.

  3. Q6. At 731 K, Kc = 46.4 for H2 + I2 ⇌ 2HI. A mixture has [H2] = [I2] = 0.10 mol L−1 and [HI] = 0.50 mol L−1. Is it at equilibrium?

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    [HI]2/[H2][I2] = 0.25/0.010 = 25, not 46.4, so it is not at equilibrium. The ratio is too small, so more HI must form before equilibrium is reached.

  4. Q7. At 731 K, Kc = 46.4 for H2 + I2 ⇌ 2HI. Find the constant for HI ⇌ ½H2 + ½I2.

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    This is the halved equation reversed: 1/(46.4)1/2 = 1/6.81, about 0.147.

  5. Q8. At a certain temperature, K1 = 2.0 for A ⇌ B and K2 = 0.20 for B ⇌ 2C. Find K for A ⇌ 2C and for 2C ⇌ A.

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    Adding the equations multiplies the constants: K = 2.0 × 0.20 = 0.40 for A ⇌ 2C. Its reverse, 2C ⇌ A, has 1/0.40 = 2.5.