If 16.9 kJ are released when 1.50 g of O2 reacts with an excess of NO, complete the thermochemical equation below.
4 NO(g) + 3 O2(g) 2 N2O5(g)
ÄHrxn=???
Wouldn't that be 16.9 kJ x (32/1.5) x 3 = ??
I worked the problem out and got 1081.6. I need the least number of significant digits..so how would i round it?
To determine the standard enthalpy change (∆Hrxn) for the given thermochemical equation, we need to use the given information about the energy released (16.9 kJ) when 1.50 g of O2 reacts with an excess of NO.
The balanced equation is:
4 NO(g) + 3 O2(g) -> 2 N2O5(g)
Let's break down the steps to determine the ∆Hrxn:
Step 1: Calculate the molar mass of O2
The molar mass of O2 is 32 g/mol since oxygen's atomic mass is 16 g/mol and there are two oxygen atoms in one O2 molecule.
Step 2: Calculate the moles of O2 reacted
To calculate the moles of O2 reacted, divide the given mass (1.50 g) by the molar mass of O2:
moles = mass / molar mass
moles = 1.50 g / 32 g/mol
Step 3: Determine the moles of NO reacted
According to the balanced equation, the stoichiometry between O2 and NO is 3:4. Therefore, the moles of NO reacted will be (3/4) times the moles of O2 reacted.
Step 4: Calculate the energy released per mole of O2 reacted
To determine the energy released per mole of O2 reacted, divide the given energy (16.9 kJ) by the moles of O2 reacted.
Step 5: Calculate the ∆Hrxn for the balanced equation
Since the stoichiometry between O2 and ∆Hrxn is 3:2 (from the balanced equation), multiply the moles of O2 reacted by this ratio to determine the ∆Hrxn.
Following these steps, you should be able to calculate the ∆Hrxn for the balanced equation using the given information.