Nitrogen oxide (NO) has been found to be a key component in the many biological processes. It also can react with oxygen to give the brown gas NO2. When one mole of NO reacts with oxygen, 57.0 kJ of heat is evolved.
a) Write the thermochemical equation for the reaction between one mole of nitrogen oxide and oxygen.
Is a thermochemical equation just like any other equation? I have written:
2NO + O2 --> 2NO2
b)Is the reaction exothermic or endothermic?
I know what endothermic and exothermic reactions are, but how do I determine this from the information given?
d) What is delta-H when 5.00 g of NO reacts?
What I had is this:
5.00 g x 1 mol/30.01 g = .1666 mol
1/.1666= 6.00
delta-H= q
q= 57.0 kj/mol / 6.00 = 9.50 kJ
I'm not sure if this makes sense or not.
e) How many grams of nitrogen oxide must react with an excess of oxygen to liberate ten kilojoules of heat?
This sounds a bit like a limiting-excess reaction to me, but I'm not sure if I need to use that to complete the problem. So far I have written
10.00 kJ = m * Cp * delta-T
However this leaves me with two unknowns, so that's not a big help.
a) Yes and no. A thermo equation is just like any other equation EXCEPT that they are written for 1 mol of the material. Besides, the problem states to use 1 mol NO.
NO + 1/2 O2 ==> NO2
Thermochemists aren't bothered by the 1/2 O2.
b)The problem states that 57.0 kJ of energy is EVOLVED (hint: given off?). So that makes it an __________ reaction.
c)You are correct with the number but is the sign correct? If heat is given off then delta H should be negative?
d. 30.01 g NO will react to give 57.0 kJ of heat. So how many g NO are required to give 10 kJ of heat?
30.01 g NO x 10 kJ/57.0 kJ = ??
9.50kJ
a) Yes, your thermochemical equation is correct:
2NO + O2 -> 2NO2.
b) The reaction is exothermic because heat is evolved during the reaction.
c) Your calculations for delta-H when 5.00 g of NO reacts are correct. However, the sign of delta-H should be negative since heat is evolved during the reaction. Therefore, the correct answer is -9.50 kJ.
d) To find the grams of nitrogen oxide required to liberate ten kilojoules of heat, you can use the proportionality of mass and heat released.
Let x be the grams of NO required:
30.01 g / 57.0 kJ = x / 10 kJ
Solving for x, you get:
x = (10 kJ * 30.01 g) / 57.0 kJ = 5.263 g
Therefore, approximately 5.263 grams of nitrogen oxide must react with an excess of oxygen to liberate ten kilojoules of heat.
a) The thermochemical equation for the reaction between one mole of nitrogen oxide (NO) and oxygen (O2) is written as:
2NO + O2 -> 2NO2
Note that balancing the equation is important to ensure the conservation of atoms in the reaction.
b) To determine if the reaction is exothermic or endothermic, we need to consider the sign of the heat evolved. In this case, it is stated that 57.0 kJ of heat is evolved during the reaction. When heat is given off or released, it is an exothermic reaction.
c) The calculation for delta-H is correct. However, there is an error in the sign. Since heat is being evolved (given off), the sign of delta-H should be negative. Therefore, the correct delta-H value for the reaction of 5.00 g of NO is -9.50 kJ.
d) To determine the number of grams of nitrogen oxide (NO) that must react with an excess of oxygen to liberate ten kilojoules (10.00 kJ) of heat, we can use the given information about the heat evolved in the reaction.
One mole of NO (30.01 g) reacts to give 57.0 kJ of heat. We can set up a proportion to find the mass of NO required:
30.01 g NO / 57.0 kJ = x g NO / 10.00 kJ
Simplifying the proportion, we find:
x = (30.01 g NO * 10.00 kJ) / 57.0 kJ
Calculate the value of x to find the mass of NO required to produce 10.00 kJ of heat.
Note: It is not necessary to use the specific heat capacity (Cp) and temperature change (delta-T) in this problem, as the heat evolved is directly provided in the question.