How much heat is absorbed/released when 35.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) according to the following chemical equation?
4NH3(g)+5O2(g)--->4NO(g)+6H2O(l)
DeltaH=1168kj
yeah it just says rleased/absorbed i guess that's what they want you to find out im not sure what you mean by 1168 kJ/mole or 1168 kJ/reaction?
Is that 1168 kJ for the reaction (4 moles NH3 or 68 grams) or kJ/mol (for 17 grams).
428.6kJ of heat
To calculate the amount of heat released or absorbed in a chemical reaction, you need to use the equation:
q = m * C * ΔT
Where:
q is the amount of heat absorbed or released (in Joules)
m is the mass of the substance (in grams)
C is the specific heat capacity of the substance (in J/g·°C)
ΔT is the change in temperature (in °C)
In this case, we have the enthalpy change (ΔH) given as 1168 kJ, but we need it in units of J. So let's convert it:
1 kJ = 1000 J
Therefore, ΔH = 1168 kJ = 1168 * 1000 J = 1,168,000 J
Now, let's calculate the amount of heat absorbed or released:
1. Determine the number of moles of NH3 using the given mass (35.00 g) and the molar mass of NH3.
The molar mass of NH3 is:
1(14.01) + 3(1.01) = 17.03 g/mol
Number of moles of NH3 = mass / molar mass
Number of moles of NH3 = 35.00 g / 17.03 g/mol
2. Use the stoichiometric coefficients of NH3 and ΔH to determine the amount of heat absorbed or released.
From the balanced chemical equation:
4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(l)
We can see that the stoichiometric coefficient of NH3 is 4.
Therefore, the amount of heat absorbed or released can be calculated as follows:
q = (number of moles of NH3) * (ΔH) / (stoichiometric coefficient of NH3)
q = ( moles of NH3) * ( ΔH ) / ( 4 )
Finally, plug in the calculated values to find the answer.
Is that 1168 kJ/mole or 1168 kJ/reaction?
And is the problem not telling whether it is absorbed/released on purpose?
Yes, 601 kJ is right and it is released; i.e., the 1168 is -1168 kJ for the reaction as written.
1168 x (35.0/68) = 601 kJ released or delta H = -601 kJ.