When a chemist burns ammonia according to the reaction below she finds that the reaction releases heat. (It is exothermic.)
4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)
The enthalpy of the reaction DH = -1267 kJ. What is the enthalpy change (in kJ) when 7 grams of ammonia are burned?
You need to find the arrow key and use it. Without an arrow how do we tell the reactants from the products. Fortunately I know where it is but in the future remember this tip.
dHrxn is -1267 kJ and that is for 4*17 = 68g NH3. YOu want to know dH for 7 g NH3.
That's -1267 kJ x (7/68) = ? kJ
To find the enthalpy change when 7 grams of ammonia are burned, we need to use the molar mass of ammonia (NH3) and convert grams to moles. Then, we can use the stoichiometry of the balanced equation to determine the amount of heat released.
1. Find the molar mass of ammonia (NH3):
- The molar mass of nitrogen (N) is 14.01 g/mol.
- The molar mass of hydrogen (H) is 1.01 g/mol.
- Multiply the molar mass of nitrogen by the subscript (1) and the molar mass of hydrogen by the subscript (3) in the formula of ammonia (NH3).
- Molar mass of NH3 = 14.01 g/mol + 3 * 1.01 g/mol = 17.03 g/mol.
2. Calculate the number of moles of ammonia (NH3):
- Use the equation: Moles = Mass / Molar mass.
- Moles of NH3 = 7 g / 17.03 g/mol ≈ 0.411 moles.
3. Determine the enthalpy change using the stoichiometry of the balanced equation:
- The balanced equation is: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g).
- From the balanced equation, we can see that for every 4 moles of NH3, 1267 kJ of heat is released.
- Since the balanced equation gives the relationship between moles of NH3 and heat released, we can use a proportion to find the heat released by 0.411 moles of NH3.
- (0.411 moles NH3 / 4 moles NH3) = (Heat released / 1267 kJ).
- Heat released = (0.411 moles NH3 / 4 moles NH3) * 1267 kJ ≈ -131 kJ.
Therefore, the enthalpy change when 7 grams of ammonia are burned is approximately -131 kJ.