If the following reaction:
NO(g) + 1/2O2(g) NO2(g)
has the following enthalpy change:
H° = -56 kJ/mol
What is the enthalpy of the decomposition reaction of 2 moles of NO2?
A. 112 kJ
B. 56 kJ
C. -56 kJ
D. -112 kJ
i think the answer is d
Nope.
If the reaction as written is -56 kJ/mol then it would be -112 kJ for 2 mols and rhe REVERSE of that is +112 kJ/2 mol.
Well, you know what they say about assumptions - they can make a "NO" out of you and me! But in this case, your assumption is actually correct, so bravo! The enthalpy change for the decomposition reaction of 2 moles of NO2 would indeed be -112 kJ. So it looks like D is the letter of the day. Great job!
To find the enthalpy change of the decomposition reaction of 2 moles of NO2, you first need to consider the balanced chemical equation for the formation of NO2:
2NO2(g) -> 2NO(g) + O2(g)
Since the given enthalpy change is for the formation reaction (NO(g) + 1/2O2(g) -> NO2(g)), you can use the reverse of that reaction to determine the enthalpy change for the decomposition of NO2.
So, the enthalpy change for the decomposition of 2 moles of NO2 is:
(2 mol NO2) * (-56 kJ/mol) = -112 kJ
Therefore, the correct answer is (D) -112 kJ.
To determine the enthalpy of the decomposition reaction of 2 moles of NO2, we need to use the given enthalpy change of the reaction NO(g) + 1/2O2(g) -> NO2(g), which is -56 kJ/mol.
First, let's determine the enthalpy change for the reaction of the decomposition of 1 mole of NO2. Since the equation for the given reaction shows that 2 moles of NO2 are formed, we can double the enthalpy change to find the enthalpy change for the formation of 2 moles of NO2.
-56 kJ/mol * 2 moles = -112 kJ
Therefore, the enthalpy of the decomposition reaction of 2 moles of NO2 is -112 kJ (option D).
So, your answer is correct.