Consider the reaction N2+O2-->2NO with Delta H=43.2kcal.
When 50.0g of N2 react with an excess of oxygen, how much heat is generated/consumed?
I know that this is an endothermic reaction b/c of the + delta H, is there some sort of formula to use here? I have no idea where to start...can someone walk me through this? I'm lost!!!
Yes, it's endothermic and you're right for the reason, too.
The equation tells you that 43.2 kcal is consumed by 1 mol N2 (28g). All you need to do is proportion that out for 50 g.
43.2 kcal x (50/28) = ?
Well, well, well, it seems like you're in quite a pickle! Don't worry, I'm here to help you. But before we dive into the depths of this problem, let me just say, chemistry can be quite exothermic... I mean, exciting!
Now, let's get down to business. To determine how much heat is generated or consumed in a reaction, we can use the good old formula:
q = m * ΔH
Where:
q is the amount of heat (in joules or calories, depending on the units of ΔH)
m is the mass of the substance involved in the reaction (in grams)
ΔH is the enthalpy change (in joules or calories)
In this case, we are given that ΔH is 43.2 kcal (which is the same as 43.2 x 10^3 cal). But hold your horses, we need to convert grams to moles first.
To convert 50.0 grams of N2 to moles, we use the molar mass of N2, which is approximately 28 g/mol. So, we do a little math:
50.0 g N2 * (1 mol N2 / 28 g N2) = 1.79 mol N2
Now, we can plug these values into the formula:
q = (1.79 mol N2) * (43.2 x 10^3 cal/mol N2)
You might notice that I multiplied by the molar enthalpy (ΔH) in calories, since the mass was given in grams. If the mass were given in kilograms, we would multiply by the molar enthalpy in joules.
Anyway, let's calculate this:
q ≈ 77.4 x 10^3 cal
So, approximately 77.4 kilocalories of heat are generated or consumed in this reaction. Remember, since the enthalpy change is positive, it means that this reaction is endothermic. It'll just hog all the heat!
I hope this helps you out, my friend. And remember, if chemistry gets you down, just remember that laughter is the best catalyst!
To determine the amount of heat generated or consumed in a chemical reaction, you can use the formula:
q = m * ΔH
where:
q is the heat change (in calories or joules)
m is the mass of the substance involved in the reaction (in grams)
ΔH is the molar heat of the reaction (in calories or joules per mole)
In this case, we need to find out the heat change when 50.0g of N2 reacts.
First, we need to calculate the number of moles of N2 present. We can use the molar mass of N2, which is approximately 28 g/mol, to determine this.
moles of N2 = mass of N2 / molar mass of N2
moles of N2 = 50.0g / 28 g/mol
Next, we can use the balanced equation to determine the molar heat of the reaction. The coefficient of N2 in the balanced equation is 1, so the molar heat of the reaction is 43.2 kcal/1 mol of N2.
Finally, we can calculate the heat change using the formula:
q = m * ΔH
Substituting the known values:
q = (moles of N2) * ΔH
q = (50.0g / 28 g/mol) * 43.2 kcal
Now, you can calculate the exact amount of heat generated or consumed in the reaction by substituting the values into the equation and performing the calculations.
To determine the amount of heat generated or consumed in a chemical reaction, you can use the formula:
q = m·ΔH
where:
q is the heat generated or consumed in the reaction
m is the mass of the reactant being used or produced
ΔH is the enthalpy change of the reaction
In this case, you have 50.0g of N2 and the enthalpy change (ΔH) is given as 43.2 kcal. Since N2 is the reactant, its mass will be used in the calculation.
First, let's convert the mass of N2 to moles. To do this, we need to know the molar mass of N2, which is approximately 28 g/mol. So, the number of moles can be calculated as:
moles of N2 = mass of N2 / molar mass of N2
= 50.0 g / 28 g/mol
≈ 1.79 mol
Now that we have the moles of N2, we can calculate the amount of heat generated or consumed by the reaction using the formula mentioned earlier:
q = m·ΔH
Substituting the known values:
q = (1.79 mol) · (43.2 kcal/mol)
≈ 77.33 kcal
Therefore, when 50.0g of N2 reacts with an excess of oxygen in this reaction, approximately 77.33 kcal of heat are generated.