If enthalpies of formation of C2H4( g),CO2( g) and H2O(l) at 25∘C and 1 atm pressure are 52,394 and −286 kJ/mol respectively, the change in enthalpy for combustion of C2H4 is equal to

Which reaction shows that the enthalpy of formation of C2H4 is Hf = 52.5 kJ/mol?A.2C(s) + 2H2(g) + 52.5 kJ C2H4B.2C(s) + 4H(g) C2H4 + 52.5 kJC.2C(s) + 2H2(g) C2H4 + 52.5 kJD.2C(s) + H(g) + 52.5 kJ C2H4SUBMITarrow_backPREVIOUS

Find the enthalpy of formation of acetylene (C2H2): 2C+H2  C2H2If the enthalpy of combustion of C, H2 and C2H2 are -400kj.mol-1, -300kj.mol-1 and-1300kj.mol-1?

At constant pressure, the combustion of 5.00 g of C2H6(g) releases 259 kJ of heat. What is ΔH for the reaction given below? 2 C2H6(g) + 7 O2(g)–> 4 CO2(g) + 6 H2O(l). Hint: Convert 5.0 grams to moles (divide by molar mass).  Divide 259 kJ by your answer to get kJ/mole reacted.  Next, multiply by 2 since there are 2 C2H6(g) in the balanced reaction. Group of answer choices2155 kJ1800 kJ3108 kJ5168 kJ

Given the following thermochemical data, what is the enthalpy of formation ΔfHo for C2H5OH(l) C(s)  +  O2(g)  →  CO2(g)                               ΔHo = –393 kJ mol–1H2(g)  +  1/2 O2(g)    →  H2O (g)                  ΔHo = -298 kJ mol–1C2H5OH +3O2 →  3H2O (g) + 2CO2(g)       ΔHo =–1369 kJ mol–1 Use-full formula ΔrxnHo=  ∑ ΔfHoproducts -  ∑ ΔfHoreactants

Using Hess’s Law, determine the enthalpy of reaction for,C2H4(g)   +  H2(g)   –>  C2H6(g)     ΔH = ?Using the following reactions:C2H4(g)   +   3O2(g)   –>  2CO2(g)   +  2H2O(l)  ΔH = -1401 kJC2H6(g)  +  7/2O2(g)   –>   2CO2(g)   +  3H2O(l)  ΔH = -1550 kJH2(g)  +    1/2O2(g)  –>  H2O(l)    ΔH = -286 kJGroup of answer choices243 kJ-311 kJ95 kJ-137 kJ