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Answer :
Using the given enthalpy change and stoichiometry, the heat transfer for 1.30 mol of [tex]\( H_2 \)[/tex] is calculated. The correct answer is E.) 315 kJ released during the reaction.
To determine the heat transfer when 1.30 mol of [tex]\( H_2 \)[/tex] reacts, we need to use the given enthalpy change [tex](\( \Delta H_{rxn} = -484 \, \text{kJ} \))[/tex] and the stoichiometry of the reaction.
Given that the reaction [tex]\( 2H_2 + O_2 \rightarrow 2H_2O \)[/tex] has a molar coefficient of 2 for [tex]\( H_2 \)[/tex], we can infer that the enthalpy change corresponds to the reaction of 2 moles of [tex]\( H_2 \)[/tex]. Thus, to find the heat transfer for 1.30 mol of [tex]\( H_2 \),[/tex] we'll use a proportion:
[tex]\[ \frac{-484 \, \text{kJ}}{2 \, \text{mol} \, H_2} = \frac{x \, \text{kJ}}{1.30 \, \text{mol} \, H_2} \][/tex]
Solving for [tex]\( x \)[/tex], we find:
[tex]\[ x = \frac{-484 \, \text{kJ} \times 1.30 \, \text{mol} \, H_2}{2 \, \text{mol} \, H_2} = -314.6 \, \text{kJ} \][/tex]
Since the negative sign indicates that heat is released (exothermic reaction), the correct answer is:
E.) 315 kJ released
The question probable maybe:
Consider the reaction 2H_2+O_2⟶2H_2O Δ_{rxn}=−484 kJ Which answer best describes the transfer of heat that occurs when 1.30 mol H_2 reacts?
A.) 484 kJ absorbed
B.) 629 kJ released
C.) 484 kJ released
D.) 629 kJ absorbed
E.) 315 kJ released
F.) 315 kJ absorbed
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