Question

A student performs a neutralization reaction involving an acid and a base in an open polystyrene coffee-cup calorimeter. how would the calculated value of dh differ from the actual value if there was significant heat loss to the surroundings? group of answer choices

-dh calc would be negative, but less negative than the actual value (whether that actual value is positive or negative).
-dh calc would be positive, but more positive than the actual value (whether that actual value is positive or negative).
-dh calc would be positive, but less positive than the actual value (whether that actual value is positive or negative).
-dh calc would be negative, but more negative than the actual value (whether that actual value is positive or negative).

Answer 1

The calculated value of ΔH in a neutralization reaction would be positive, but less positive than the actual value if there was significant heat loss to the surroundings.

Step-by-step explanation:

The calculated value of ΔH (enthalpy change) in a neutralization reaction would differ from the actual value if there was significant heat loss to the surroundings. The correct option would be: -ΔH calc would be positive, but less positive than the actual value (whether that actual value is positive or negative).

When there is significant heat loss to the surroundings, the measured value of ΔH calculated using a calorimeter will be lower than the actual value of ΔH for the reaction. This is because the lost heat is not accounted for in the calculation, resulting in a lower value. However, it should be noted that the sign of ΔH (positive or negative) is not affected by the heat loss to the surroundings.