Question

What is the enthalpy change when 41.8 mL of 0.530 M sulfuric acid reacts with 21.5 mL of 0.175 M potassium hydroxide?

a) ΔH = -111.6 kJ/mol
b) ΔH = (-111.6 kJ/mol) * Number of moles reacted
c) Insufficient information provided.
d) ΔH = (-111.6 kJ/mol) / 41.8 mL * 21.5 mL

Answer 1

The enthalpy change can be calculated using the formula ΔH = q / (mol Acid). First, determine the number of moles of acid and base reacted. Then, calculate the enthalpy change using the given value of q and the moles of acid.

Step-by-step explanation:

The enthalpy change can be calculated using the formula:

ΔH = q / (mol Acid)

First, we need to determine the number of moles of acid (sulfuric acid) and base (potassium hydroxide) reacted:

Moles of acid = Volume of acid (L) x Concentration of acid (M)

Moles of base = Volume of base (L) x Concentration of base (M)

Next, we calculate the enthalpy change using the given value of q and the moles of acid:

ΔH = q / Moles of acid

Given the information provided, we can calculate the enthalpy change for the reaction.