Hydrazine (n2h4) is a fuel used by some spacecraft. it is normally oxidized by n2o4 according to the following equation:
n2h4(l)+n2o4(g)-> 2n2o(g)+2h2o(g)

calculate delta hrxn for this reaction using standard enthalpies of formation.

-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.

b. examine and describe cells