If 17.7 kJ are released when 1.00 g of O2 reacts with an excess of NO, complete the thermochemical equation below.

4 NO(g) + 3O2(g)=2 N2O5(g) ΔHrxn = ___KJ

To find the enthalpy change (ΔH) for the given thermochemical equation, we can use the given information that 17.7 kJ are released when 1.00 g of O2 reacts.

First, we need to determine the molar mass of O2. The molar mass of O2 is calculated by adding together the atomic masses of oxygen, which is 16.00 g/mol.

Next, we can convert the mass of O2 (1.00 g) into moles using the molar mass of O2.

moles of O2 = mass of O2 / molar mass of O2
= 1.00 g / 32.00 g/mol
= 0.03125 mol

From the balanced equation, we can see that 3 moles of O2 react to form 2 moles of N2O5.

Therefore, the amount of energy released per mole of O2 can be determined by dividing the given energy change (17.7 kJ) by the number of moles of O2.

Energy change per mole of O2 = 17.7 kJ / 0.03125 mol
= 566.4 kJ/mol

Since this value represents the energy change per mole of O2, we need to multiply it by the stoichiometric coefficient of O2 from the balanced equation (3).

ΔHrxn = Energy change per mole of O2 × Stoichiometric coefficient of O2
= 566.4 kJ/mol × 3
= 1699.2 kJ/mol

Thus, the enthalpy change (ΔH) for the given thermochemical equation is 1699.2 kJ.

To solve this thermochemical equation, we need to determine the heat of reaction (ΔHrxn) in kJ.

Given that 17.7 kJ is released when 1.00 g of O2 reacts, we can calculate the heat released per gram of O2.

First, we need to determine the molar mass of O2, which is 32.00 g/mol.

Next, we convert the mass of O2 to moles using the equation:

moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 1.00 g / 32.00 g/mol
moles of O2 = 0.03125 mol

Since the stoichiometric coefficient of O2 in the balanced equation is 3 (meaning it reacts in a 3:3 ratio with NO), we can calculate the heat of reaction per mole of O2:

ΔHrxn (per mole of O2) = 17.7 kJ / 0.03125 mol
ΔHrxn (per mole of O2) ≈ 566.4 kJ/mol

Finally, since the stoichiometric coefficient of O2 is 3 in the balanced equation, we multiply the ΔHrxn (per mole of O2) by 3:

ΔHrxn = 566.4 kJ/mol * 3
ΔHrxn = 1699.2 kJ

Therefore, the ΔHrxn for the given thermochemical equation is 1699.2 kJ.

1 mol NO = 14 + 16 = 30 grams.

4 mol NO = 120 g
17.7 kJ are released for 120 g NO so how much will be released for 1 g? That's
17.7 kJ x (1/120) = ? kJ. Released kJ means an exothermic reaction and that makes the dH = a negative number.