Part A Calculate the enthalpy change, delta H,for the expansion of methane CH4(g

Question

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 daa 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.

Explanation / Answer

A)
delta H = 4*delta Hf(H(g)) + delta Hf(C(g)) - delta Hf(CH4(g))
     = 4*(217.94) + (718.4) - (-74.8)
      =1664.96 KJ/mol

B)
delta H = 2*delta Hf(H2(g)) + delta Hf(C(S)) - delta Hf(CH4(g))
     = 2*(0) + (0) - (-74.8)
      = 74.8 KJ/mol

C)
CH4(g) + 4F2(g) ------> CF4 (g) + 4HF (g)
delta H = 4*delta Hf(HF(g)) + delta Hf(CF4(g)) - delta Hf(CH4(g)) - 4*delta Hf(F2(g))
delta H = 4*(-268.61) + (-679.9) - (-74.8) - 4*(0)
                = -1679.54 KJ/mol

when 0.570 mol of methane CH4(g) is reacted with 0.720 mol of fluorine,
F2 is limiting reagent
So we will use F2 in our futher calculation
4 mol of F2 releases 1679.54 KJ
0.72 mol will release = 1679.54*0.72/4 = 302.32 KJ
Answer: 302.32 KJ

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