I originally used 1 gram of NaOH. What if I used 2 grams of NaOH. Would that change the heat evolved, Q and change in ethalpy, (delta)H?
Through direct calculation, I see that the Q change by only about 0.1 kcal, but for (delta)H, I see that the value halved(which is expected since the mole of NaOH would be doubled).
However, this is based on the assumption that the change in temperature remains the same. I'm not sure if the temperature will be smaller or larger with an increase in temperature.
Can you help me with this problem? Thanks!
You need to clarify your question. I assume you have done something such as adding NaOH to water and measuring the temperature. Adding twice the mass of NaOH WILL change Q.
Yes, NaOH is added to the water. How much will adding twice the mass of NaOH will change Q? I am asked what would have been the heat evolved based on my actual observations and what effect would this have on my calculation of (delta)H.
Would saying "based on the assumption that the change in temperature remains the same, the Q change by only about 0.1 kcal, but for (delta)H, I see that the value halved(which is expected since the mole of NaOH would be doubled)" be sufficient? I am not sure how much the temperature will change by adding twice the mass.
Thanks for the help!
Adding twice the amount of NaOH to the same amount of water will double Q but delta H per gram or delta H per mole will not change. If you ASSUME that the temperature will not change (a false assumption) then delta H will be halved. From a practical standpoint, does it make sense that if 1 g NaOH in 100 mL water changes the temperature of the water by 10 degrees that adding 2 grams would also change the temperature of the same water by 10 degrees. Doesn't sound reliable to me.
Certainly! I can assist you with this problem.
When calculating the change in heat, Q, and the change in enthalpy, ΔH, for a chemical reaction, it is important to consider the stoichiometry of the reaction. Let's break down the steps to find these values.
1. Determine the balanced chemical equation:
You haven't provided the specific reaction you are working with, but for the sake of illustration, let's consider the reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
2. Determine the molar ratio:
According to the balanced equation, the molar ratio of NaOH to HCl is 1:1. This means that for every 1 mole of NaOH used, 1 mole of HCl is consumed.
3. Calculate the moles of NaOH:
Since you mentioned using 1 gram of NaOH initially, we need to convert this mass to moles. The molar mass of NaOH is about 40 g/mol, so:
moles of NaOH = mass / molar mass
moles of NaOH = 1 g / 40 g/mol
moles of NaOH = 0.025 mol
4. Calculate the moles of HCl:
Since the molar ratio is 1:1, the moles of HCl is also 0.025 mol.
5. Calculate the heat evolved, Q:
The heat evolved can be determined using the equation:
Q = moles of HCl × ΔH (per mole of HCl)
Now, if you double the amount of NaOH to 2 grams, the new moles of NaOH would be:
moles of NaOH = 2 g / 40 g/mol
moles of NaOH = 0.05 mol
Since the molar ratio is still 1:1, the moles of HCl would also be 0.05 mol.
To find the change in heat, Q, we can use the equation:
Q = moles of HCl × ΔH (per mole of HCl)
Since the moles of HCl have doubled, the value of Q will also double. So, by doubling the amount of NaOH, the heat evolved, Q, would increase by approximately 0.1 kcal. This matches your observation.
Now, for the change in enthalpy, ΔH, the value halves because you're comparing the same reaction with two different amounts of NaOH. The change in enthalpy is an extensive property, meaning it depends on the quantity of substances involved in the reaction. Thus, when you double the moles of NaOH, the value of ΔH will halve.
Regarding temperature, it's important to note that the change in temperature is not affected by the amount of NaOH used. The change in temperature is a result of the chemical reaction and the heat evolved, not the quantity of reactants. So, regardless of the amount of NaOH used, the change in temperature should remain the same.
I hope this explanation helps! Let me know if you have any further questions.