When 1.92 g of magnesium reacts with nitrogen to form magnesium nitride, the hea
ID: 1025134 • Letter: W
Question
When 1.92 g of magnesium reacts with nitrogen to form magnesium nitride, the heat evolved is 12.2 kJ. Calculate the standard enthalpy of formation of Mg.N, V. VI. Use the thermodynamic data from Appendix IV in your textbook to calculate the standard enthalpy change for the following reaction at 25 25O, (g) + O2 (g) 2SO, (g) What is the enthalpy change for this reaction if 10.0 g of sulfur dioxide reacted? VII. Calculate the work (in J) that must be done by the reaction at STP (0 °C and 1 atm) to make room for the gaseous products of octane combustion when 2.00 mol of octane is burned. Hint The reaction Shwat 25 mol fa mc b)are replaced by 34 mo (Hint: The reaction shows that 25 mol of gas molecules (reactants) are replaced by 34 mol of gas molecules (products), a net increase of 9 mol. Because the molar volume of a gas at STP is 22.4 L, the change in volume (AV) and hence the work needed for the volume expansion of the gases at 1 atm can be calculated.)Explanation / Answer
3Mg+ N2 ------->Mg3N2, Magnesium nitride
molar mass of Mg= 24, moles of Mg given = mass/molar mass= 1.92/24=0.08
3 moles of Mg gives rise to 12.2 Kj of heat
0.08 moles of Mg gives rise to 12.*0.08/3=0.325 Kj of hear
2. for the reaction 2SO2(g)+ O2(g) -------->2SO3,
deltaHo= standard enthalpy of reaction =2* standard enthalpy change of SO3- (2* standard enthalpy change of SO2+1* standard enthalpy change of O2)
standard enthalpy data (Kj/mole) : SO2= -297, SO3= -396, O2=0
deltaHo= 2*(-396)- {2*-297+1*0}=-198 Kj, 2,2 and 1 are coefficients of SO2, SO3 and O2 respectively in the reaction
2 mole of SO2 correspond to 2*(32+32)= 128 gm of SO2
128 gm of SO2 produces 198 kj of heat
10 gm produces 10*198/128 gm of Heat=15.5 KJ of heat
the enthalpy change of reaction =-15.5 Kj
3. molar volume of gases at STP= 22.4 L/mole
there is change in moles to the tune of 9. Hence change in volume= 22.4 L/mole*9= 201.6 L
work done = P*change in volume = 1 atm* 201.6 L =201.6 L.atm
1L. atm= 101.3 joules
work done = 201.6*101.3 joules =20422 joules
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