When 1 mol of NO(g) forms from its elements, 90.29 kJ of heat is absorbed, How much heat is involved when 4.29 g of NO decomposes to its elements?
is it -12.90 kJ ???
thank you.
I obtained -12.91 kJ using 30 for the molar mass of NO.
To solve this problem, we need to use the concept of molar mass and stoichiometry. Here's the step-by-step explanation:
1. Start by calculating the number of moles of NO(g) in 4.29 g of NO. To do that, divide the given mass by the molar mass of NO. The molar mass of NO is calculated as follows:
Molar mass of N = 14.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of NO = Molar mass of N + (Molar mass of O × 1)
= 14.01 + (16.00 × 1) g/mol
= 30.01 g/mol
Now, divide the given mass (4.29 g) by the molar mass (30.01 g/mol) to obtain the number of moles:
Moles of NO = Mass of NO / Molar mass of NO
= 4.29 g / 30.01 g/mol
= 0.143 moles
2. Next, use the balanced chemical equation to determine the stoichiometric relationship between the number of moles of NO and the amount of heat involved. The balanced equation for the formation of NO from its elements is:
N2(g) + O2(g) -> 2NO(g)
According to this equation, 1 mole of NO is formed by reacting 1 mole of N2 and 1 mole of O2. Therefore, the heat involved in the formation of 1 mole of NO corresponds to 90.29 kJ.
3. Finally, we can use the proportionality between moles and heat to determine the amount of heat involved when 0.143 moles of NO decompose.
To do that, we set up the following proportion:
(0.143 moles / 1 mole) = (x kJ / 90.29 kJ)
Cross-multiplying and solving for x, we get:
x = (0.143 moles / 1 mole) * 90.29 kJ
= 12.91 kJ
Thus, the amount of heat involved when 4.29 g of NO decomposes to its elements is approximately 12.91 kJ. Therefore, the answer is not -12.90 kJ, but rather 12.91 kJ.