Question

Given the reactions (1) and (2) below, determine (i) ?rH° and ?rU° for reaction (3), (ii) ?fH° for both HI(g) and H2O(g), all at 298 K. (1) H2(g) + I2(s) ? 2 HI(g) ?rH° = +52.96 kJ mol-1 (2) 2 H2(g) + O2(g) ? 2 H2O(g) ?rH° = -483.64 kJ mol-1 (3) 4 HI(g) + O2(g) ? 2 I2(s) + 2 H2O(g)

          Given the reactions (1) and (2) below, determine (i) ?rH° and ?rU° for reaction (3), (ii) ?fH° for both HI(g) and H2O(g), all at 298 K.

(1) H2(g) + I2(s) ? 2 HI(g) ?rH° = +52.96 kJ mol-1
(2) 2 H2(g) + O2(g) ? 2 H2O(g) ?rH° = -483.64 kJ mol-1
(3) 4 HI(g) + O2(g) ? 2 I2(s) + 2 H2O(g)

Given the reactions (1) and (2) below, determine (i) ?rH° and ?rU° for reaction (3), (ii) ?fH° for both HI(g) and H2O(g), all at 298 K.

(1) H2(g) + I2(s) ? 2 HI(g) ?rH° = +52.96 kJ mol-1
(2) 2 H2(g) + O2(g) ? 2 H2O(g) ?rH° = -483.64 kJ mol-1
(3) 4 HI(g) + O2(g) ? 2 I2(s) + 2 H2O(g)

Added by Sofia M.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Narayan Hari

03/31/2025

Step 1

To do this, we can multiply reaction (1) by 2 and reverse reaction (2): (1') 2H2(g) + 2I2(s) = 4HI(g) ΔH1' = 2 * (+52.96 kJ/mol) = +105.92 kJ/mol (2') 2H2O(g) = 2H2(g) + O2(g) ΔH2' = -(-483.64 kJ/mol) = +483.64 kJ/mol Now, we can add reactions (1') and (2') Show more…

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Given the reactions (1) and (2) below, determine (i) ΔrH° and ΔrU° for reaction (3), (ii) ΔfH° for both HI(g) and H2O(g), all at 298 K. (1) H2(g) + I2(s) → 2 HI(g) ΔrH° = +52.96 kJ mol-1 (2) 2 H2(g) + O2(g) → 2 H2O(g) ΔrH° = -483.64 kJ mol-1 (3) 4 HI(g) + O2(g) → 2 I2(s) + 2 H2O(g)