Answer:
-476.95 Kj
Explanation:
N2H4(l) + N2O4(g) = 2N2O(g) + 2H20(g)
∆Hrxn = n∆Hf(products) - m∆Hf(reactants)
Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation, ∆Hf = standard enthalpy of formation, ∆Hrxn= standard enthalpy of reaction.
Using the following standard enthalpies of formation ( you did not provide any ):
N2H4(l) = +50.63Kj/mol; N2O4(g) = +9.08Kj/mol; N2O(g) =+33.18Kj/mol; H2O(g) = -241.8Kj/mol
∆Hrxn = [ (2(∆Hf(N2O)) + (2(∆Hf(H2O))] – [(1(∆Hf(N2H4)) + (1(∆Hf(N2O4))]
∆Hrxn = [ 2(+33.18) + 2(-241.8)] – [ (+50.63) + (+9.08)]
∆Hrxn = [ (+66.36)+(-483.6)] – [ +50.63+9.08]
∆Hrxn = [ +66.36-483.6] – [+59.71]
∆Hrxn = -417.24-59.71
∆Hrxn = -476.95 Kj
NOTE: Remember to use the standard enthalpies of formation given to you by your instructor if they differ from the values used herein, and follow the same procedure.
