Question

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide gas (HBr). The reaction releases 72.58 kJ of heat. Which equation is the correct form to show the standard molar enthalpy of formation?

H2(g) + Br2(l) → 2HBr(g) ΔHf0= -145.16 kJ

A.1/2H2(g) + Br2(l) → HBr(g) ΔHf0= -72.58 kJ

B.H2(g) + Br2(l)→ HBr(g) ΔHf0= -36.29 kJ

C.1/2 H2(g) + Br2(l) → HBr(g) ΔHf0= -72.58 kJ

D.1/2H2(g) + Br2(l) → HBr(g)ΔHf0= -36.29 kJ

Answer 1

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of $H_2$ gas and 1 mole of $Br_2$ liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

$H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ$

Divide the equation by 2.

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$