Calculate ∆Hrxn for the following reaction at 298 K using the ∆Hof values given below. 2 Na(s) + 2 H2O(l) → 2 NaOH(s) + H2(g) Given: ∆Hof (NaOH(s)) = −426.8 kJ/mol ∆Of (H2O(l)) = −285.8 kJ/mol ∆Hof (H2(g)) = 0 kJ/mole ∆Hof (Na(s)) = 0 kJ/mole