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How much heat will be absorbed when 38.2 g of Bromine reacts with excess H2 according to the following equation?

\[ \text{H}_2 + \text{Br}_2 \rightarrow 2\text{HBr} \]

\[ \Delta H^\circ = 72.80 \text{kJ} \]

Answer :

Final answer:

The heat absorbed when 38.2g of Bromine reacts with excess H2 according to the given equation is approximately 17.4 kJ.

Explanation:

The enthalpy change of the reaction is given by the equation H2 + Br2 -> 2HBr. The value of the standard enthalpy change, ΔH°, is 72.80 kJ. To find the heat absorbed when 38.2g of bromine reacts with excess hydrogen gas, you need to first convert the mass of bromine to moles. Then, use the moles of bromine and the stoichiometry of the reaction to calculate the moles of heat absorbed.

First, find the moles of bromine:

Bromine (Br2) has a molar mass of 159.8 g/mol. Therefore, 38.2g of bromine is:

38.2 g * (1 mol/159.8 g) = 0.2396 mol

Since the coefficient of bromine in the balanced equation is 1, the moles of heat absorbed will be the same as the moles of bromine. Therefore, the heat absorbed when 38.2g of bromine reacts is 0.2396 mol * 72.80 kJ/mol = 17.42488 kJ. The answer is approximately 17.4 kJ.

Learn more about enthalpy change here:

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Rewritten by : Barada

Explanation:

First, let's transform 38.2 grams of Br2 into moles using is molar mass (MM of Br2 = 159.808 g/mol)

159.808 g --- 1 mol

38.2 g ---- x

x = 0.239 moles

Now, let's calculate how much heat will be absorbed:

72.80kJ --- 1 mol

x --- 0.239 moles

x = 17.4 kJ

Answer: 17.4 kJ