Part A Calculate the enthalpy change, delta H,for the expansion of methane CH4(g)-C(g)+4H(g) Part B Calculate the

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Part A Calculate the enthalpy change, delta H,for the expansion of methane CH4(g)->C(g)+4H(g) Part B Calculate the enthalpy change, delta H for the reverse of the formation of methane. CH4(g)->C(s) + 2H2(g) Part C Suppose that 0.570 mol of methane CH4(g) is reacted with 0.720 mol of fluorine ,F2(g),forming CF4(g) and HF (g) as sole products. Assuming that the reaction occurs at constant pressure,how much heat is released? ae aned stochiomety : Enthalpy of Reaction State and Stoichiometry Use the da?a below to answer the questions. Substance Ar C/mol) C(g) CE(G)679.9 CH (G)748 H(g) HF(g)268.61 7184 217.94 Keep in mind that the enthalpy of formation of an element in its standard state is zero.

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Part A) CH4(g) C(g) + 4H(g) Horxn = Hoproducts - Horeactants = [H(Cg) + 4 x H(Hg)] – [H(CH4g)] = [1 x 718.4 + 4 x 217.94] –[-74.8] = 1664.96 kJ/mol Part B) Hrxn = 2 x Ho(H2(g)) + Ho(C(S)) - Ho(CH4(g) = 2 x 0 + 0 – [-74.8] = 74.8 kJ/mol Part C) CH4 + 4F2 CF4 + 4HF Here, CH4 is in excess and F2 is limiting reagent From the balanced reaction Moles of CH4 reacted = 0.18 Moles of HF = 4 x moles of CH4 reacted = 4 x 0.18 = 0.72 Hrxn = Ho(CF4) + 4 x Ho(HF) – Ho(CH4) - 4 x Ho (F2) = -679.9 - 4 x 268.61 + 74.8 -0 Hrxn = -1679.54 kJ This is for 0.18 mole of CH4 reacted Therefore, Hrxn = -1679.54 x 0.18 = -302.31 kJ

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Part A Calculate the enthalpy change, delta H,for the expansion of methane CH4(g)->C(g)+4H(g) Part B Calculate the enthalpy change, delta H for the reverse of the formation of methane

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