25 cm of 1.0 mol dm− NaOH is added to 25 cm of 1.0 mol dm− HCl. The temperature rise is 6¢XC.
Which reactants will also give a temperature rise of 6¢XC?
A. 25 cm of 2.0 mol dm− NaOH and 25 cm of 2.0 mol dm− HCl.
B. 50 cm of 1.0 mol dm− NaOH and 50 cm of 1.0 mol dm− HCl.
C. 50 cm of 0.5 mol dm− NaOH and 50 cm of 0.5 mol dm− HCl.
D. 100 cm of 0.25 mol dm− NaOH and 100 cm of 0.25 mol dm− HCl.
b
To determine which reactants will also give a temperature rise of 6°C, we need to consider the stoichiometry of the reaction. The balanced equation for the reaction between NaOH and HCl is:
NaOH + HCl → NaCl + H2O
From the balanced equation, we can see that one mole of NaOH reacts with one mole of HCl. Therefore, the number of moles of the reactants must be equal in order to achieve the same temperature rise.
Let's calculate the number of moles of the reactants in each option:
A. 25 cm of 2.0 mol dm− NaOH and 25 cm of 2.0 mol dm− HCl:
Number of moles of NaOH = [(2.0 mol dm−)(25 cm)] / 1000 cm^3 = 0.05 moles
Number of moles of HCl = [(2.0 mol dm−)(25 cm)] / 1000 cm^3 = 0.05 moles
B. 50 cm of 1.0 mol dm− NaOH and 50 cm of 1.0 mol dm− HCl:
Number of moles of NaOH = [(1.0 mol dm−)(50 cm)] / 1000 cm^3 = 0.05 moles
Number of moles of HCl = [(1.0 mol dm−)(50 cm)] / 1000 cm^3 = 0.05 moles
C. 50 cm of 0.5 mol dm− NaOH and 50 cm of 0.5 mol dm− HCl:
Number of moles of NaOH = [(0.5 mol dm−)(50 cm)] / 1000 cm^3 = 0.025 moles
Number of moles of HCl = [(0.5 mol dm−)(50 cm)] / 1000 cm^3 = 0.025 moles
D. 100 cm of 0.25 mol dm− NaOH and 100 cm of 0.25 mol dm− HCl:
Number of moles of NaOH = [(0.25 mol dm−)(100 cm)] / 1000 cm^3 = 0.025 moles
Number of moles of HCl = [(0.25 mol dm−)(100 cm)] / 1000 cm^3 = 0.025 moles
Since options B, C, and D have the same number of moles of reactants as in the initial reaction, they will also give a temperature rise of 6°C. Therefore, the correct options are B, C, and D.
To determine which reactants will also give a temperature rise of 6°C, we need to consider the stoichiometry of the reaction and the amount of heat released during the reaction.
In this case, the reaction is between NaOH and HCl, and we know that 25 cm of 1.0 mol dm− NaOH and 25 cm of 1.0 mol dm− HCl result in a temperature rise of 6°C.
To solve this question, we can use the concept of "moles of heat" or "heat of reaction" (ΔH) to compare the different reactant combinations.
The "moles of heat" can be calculated using the formula:
Moles of Heat (Q) = (Volume of Solution) * (Molarity) * (ΔH)
Since the given reaction produces a temperature rise of 6°C, we can assume that the same amount of heat is released regardless of the combination of reactants.
Let's calculate the moles of heat for the given reaction:
Q = (25 cm³ + 25 cm³) * (1.0 mol dm⁻³) * (6°C)
Now, we can compare the moles of heat for each option:
Option A: Moles of Heat = (25 cm³ + 25 cm³) * (2.0 mol dm⁻³) * (6°C)
Option B: Moles of Heat = (50 cm³ + 50 cm³) * (1.0 mol dm⁻³) * (6°C)
Option C: Moles of Heat = (50 cm³ + 50 cm³) * (0.5 mol dm⁻³) * (6°C)
Option D: Moles of Heat = (100 cm³ + 100 cm³) * (0.25 mol dm⁻³) * (6°C)
We can compare the four options by comparing the moles of heat calculated. If any of the options yield the same moles of heat as the given reaction, the temperature rise would also be 6°C. So, let's calculate the moles of heat for each option.
Option A: Moles of Heat = 100 * 2.0 * 6 = 1200
Option B: Moles of Heat = 100 * 1.0 * 6 = 600
Option C: Moles of Heat = 100 * 0.5 * 6 = 300
Option D: Moles of Heat = 200 * 0.25 * 6 = 300
Now, we can see that Option C and Option D both yield the same moles of heat (300) as the given reaction. Therefore, the correct options are:
C. 50 cm of 0.5 mol dm− NaOH and 50 cm of 0.5 mol dm− HCl.
D. 100 cm of 0.25 mol dm− NaOH and 100 cm of 0.25 mol dm− HCl.
These options will also give a temperature rise of 6°C.