A calorimetry experiment determined that the combustion of 1.01 g of H2(g) reacting with excess O2(g) gave qsoln = 143 kJ. Calculate ΔH for the reaction: H2(g) + ½ O2(g) → H2O (l)

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Answer:

089

Explanation:

I have no idea

Based on the data from the calorimetry experiment, the enthalpy change, ΔH of the reaction is -286 KJ .

What is enthalpy change, ΔH of a reaction?

The enthalpy change, ΔH of a reaction is the amount of heat evolved or absorbed when reactant molecules react to form products.

From the calorimetry experiment, the determined heat of combustion of 1.01 g of H2(g) reacting with excess O2(g), qsoln = 143 kJ.

Heat is evolved in the reaction, therefore the qsoln = -143 kJ

Equation of reaction is given as:

H2(g) + ½ O2(g) → H2O (l)

Molar mass of H = 1.01 g

Mass of H2 = 1.01 g × 2 = 2.02 g

ΔH of the reaction = -143 kJ × 2.02/1.01

ΔH of the reaction = -286 KJ

Therefore, the enthalpy change, ΔH of the reaction is -286 KJ .

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