the (a) Is the reaction exothermic or endothermic. (b) Calculate the heat transf
ID: 695595 • Letter: T
Question
the (a) Is the reaction exothermic or endothermic. (b) Calculate the heat transferred when 3.55 g of Mg(s) react at constant pressure. (c) How many grams of MgO are produced during an enthalpy change of-234 kJ? (d) How many KJ of heat are absorbed when 40.3 g of MgO(s) is decomposed into Mgs) and Oz(g) at constant pressure? H =-726.5 kJ 5.4745.47) For methanol combustion: CHOH0) + 3/22 colg) + 2H00) (a) What is H for the reverse reaction? (b) If (c) Is the forward or backward reaction more likely to be favored? (d) If the reaction were written for production of H O(a), would DH be larger, smaller or the same? s are adjusted to whole numbers, what does 4H represent?Explanation / Answer
5.43) The balanced chemical equation is given.
2 Mg (s) + O2 (g) --------> 2 MgO (s); H = -1204 kJ
a) The enthalpy change, H is negative; hence, the reaction released heat and is therefore, exothermic.
b) As per the stoichiometric equation, H = -1204 kJ is the heat transferred when 2 moles of Mg react.
We have 3.55 g of Mg; the atomic mass of Mg is 24.305 g/mol. Therefore, mole(s) of Mg corresponding to 3.55 g = (3.55 g)/(24.305 g/mol) = 0.14606 mole.
The heat transferred when 0.14606 mole Mg react = (-1204 kJ)*(0.14606 mole/2 mole) = -87.92815 kJ -87.928 kJ (ans).
c) 2 moles of MgO are produced when the enthalpy change is -1204 kJ; therefore, the mole(s) of MgO produced when the enthalpy change is -234 kJ is (-234 kJ)*(2 mole)/(-1204 kJ) = 0.38870 mole.
Molar mass of MgO = (1*24.305 + 1*15.9994) g/mol = 40.3044 g/mol.
Mass of MgO produced = (0.38870 mole)*(40.3044 g/mol) = 15.66632 g 15.6663 g (ans).
d) The decomposition of MgO is given by the equation
2 MgO (s) -------> 2 Mg (s) + O2 (g)
The enthalpy change for the above reaction is the reverse of the formation process; therefore, the decomposition reaction is endothermic, i.e, absorbs heat and hence, H’ = +1204 kJ.
Mole(s) of MgO corresponding to 40.3 g = (40.3 g)/(40.3044 g/mol) = 0.99989 mole.
The heat produced is given as (+1204 kJ)*(0.99989 mole)/(2 mole) = +601.93378 kJ +601.93 kJ (ans).
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