Q1 V2The reaction between 100 mL of 0.10 M HCl and 100 mL of 0.10 M NaOH causes a temperature rise of 1.60 oC.What would be the temperature rise for a reaction of 100 mL of 0.10 M HCl with 50 ml of 0.10 M NaOH?HCl(aq) + NaOH(aq) [math: →]\ce{ \bond{-} } NaCl(aq) + H2O(l)

Question

Q1 V2The reaction between 100 mL of 0.10 M HCl and 100 mL of 0.10 M NaOH causes a temperature rise of 1.60 oC.What would be the temperature rise for a reaction of 100 mL of 0.10 M HCl with 50 ml of 0.10 M NaOH?HCl(aq) +  NaOH(aq) [math: →]\ce{ \bond{->} }  NaCl(aq) +  H2O(l)

BlackTom AI · Accepted Answer

We start by identifying the amounts of reactants involved in the two scenarios. In the given experiment, 100 mL of 0.10 M HCl provides 0.010 mol of acid, and 100 mL of 0.10 M NaOH provides 0.010 mol of base. The reaction is a neutralization, producing NaCl and H2O, and the temperature rise is tied to the amount of heat released, which scales with the extent of reaction (i.e., the number of moles that react stoichiometrically).

Option-by-option analysis:

Option a: 0.80 °C. This corresponds to half of the original heat release. If we halve the amount of NaOH to 50 mL at 0.10 M, we have 0.005 mol NaOH, which is the limiting reagent with 0.005 mol HCl remaining unreacted. Since only 0.005 mol of the acid-base neutralization occurs, the heat released should be about half of the full 0.010 mol case, giving a temperature rise of 1.60 °C × 0.5 = 0.80 °C. This matches the proportional relationship between moles reacted and temperature increase.

Option b: 3.20 °C. This would require quadrupling the heat output relative to the given case. Since you have half as many moles reacting (0.005 mol vs 0.010 mol), the temperature rise cannot exceed the original value; thus this option overestimates the result.

Option c: 1.60 °C. Keeping the same temperature rise would imply the same amount of heat release as the 0.010 mol case, but the reacting amount here is only 0.005 mol, which is half of the original neutralization extent. Therefore this choice neglects the reduced reactant quantity.

Option d: 2.40 °C. This would require more heat than the full 0.010 mol reaction produced, which is impossible since the amount of reacting species is reduced to 0.005 mol. It incorrectly assumes no change in heat per mole of reaction.

Why the other options are incorrect hinges on the direct proportionality between moles of reaction and heat released in an aqueous neutralization, assuming constant caloric conditions and complete reaction of the limiting reagent. Since halving the reacting amount halves the heat, the correct expectation is a temperature rise of 0.80 °C, not the other values.

Q1 V2The reaction between 100 mL of 0.10 M HCl and 100 mL of 0.10 M NaOH causes a temperature rise of 1.60 oC.What would be the temperature rise for a reaction of 100 mL of 0.10 M HCl with 50 ml of 0.10 M NaOH?HCl(aq) + NaOH(aq) [math: →]\ce{ \bond{->} } NaCl(aq) + H2O(l)Single choice

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