Of the following reactions, which one releases the most heat?
(1) CHig) + 20ig) - COig) + 2H20(t)
(2) 2CsHlS(t)+250ig)-1 6COig)+ 1 8H20(t)
(3) ig) - 2NO(g)
(4) 2C(s) + Hi(g) - C2Hig)
almost 5 years old and still no answers
To determine which reaction releases the most heat, we need to calculate the amount of heat released or absorbed, which is also known as the enthalpy change (∆H). The reaction with the most negative ∆H value will release the most heat.
Let's break down each reaction and calculate their enthalpy changes (∆H):
(1) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)
To calculate the ∆H for this reaction, we can use standard enthalpy of formation (∆Hf°) values. The given reaction is the combustion of methane, and the ∆Hf° values are:
∆Hf°(CO2) = -393.5 kJ/mol
∆Hf°(H2O) = -241.8 kJ/mol
∆Hf°(CH4) = -74.8 kJ/mol
Using these values, we can calculate ∆H:
∆H = ∆Hf°(CO2) + 2∆Hf°(H2O) - ∆Hf°(CH4)
= -393.5 kJ/mol + 2(-241.8 kJ/mol) - (-74.8 kJ/mol)
= -890.6 kJ/mol
(2) 2C8H18(g) + 25O2(g) -> 16CO2(g) + 18H2O(g)
Again, using ∆Hf° values:
∆Hf°(CO2) = -393.5 kJ/mol
∆Hf°(H2O) = -241.8 kJ/mol
∆Hf°(C8H18) = -249.7 kJ/mol
Calculating ∆H:
∆H = 16∆Hf°(CO2) + 18∆Hf°(H2O) - 2∆Hf°(C8H18)
= 16(-393.5 kJ/mol) + 18(-241.8 kJ/mol) - 2(-249.7 kJ/mol)
= -5101.6 kJ/mol
(3) N2(g) + 3H2(g) -> 2NO(g)
∆Hf°(NO) = 90.4 kJ/mol
∆Hf°(N2) = 0 kJ/mol
∆Hf°(H2) = 0 kJ/mol
∆H = 2∆Hf°(NO) - ∆Hf°(N2) - 3∆Hf°(H2)
= 2(90.4 kJ/mol) - 0 kJ/mol - 3(0 kJ/mol)
= 180.8 kJ/mol
(4) 2C(s) + H2(g) -> C2H2(g)
∆Hf°(C2H2) = 226.7 kJ/mol
∆Hf°(C) = 0 kJ/mol
∆Hf°(H2) = 0 kJ/mol
∆H = ∆Hf°(C2H2) - 2∆Hf°(C) - ∆Hf°(H2)
= 226.7 kJ/mol - 2(0 kJ/mol) - 0 kJ/mol
= 226.7 kJ/mol
Now, let's compare the calculated ∆H values:
∆H for reaction (1): -890.6 kJ/mol
∆H for reaction (2): -5101.6 kJ/mol
∆H for reaction (3): 180.8 kJ/mol
∆H for reaction (4): 226.7 kJ/mol
From the calculations, we can see that reaction (2) releases the most heat as it has the most negative ∆H value of -5101.6 kJ/mol.