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lakkis [162]
3 years ago
13

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g

as (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
Chemistry
1 answer:
JulsSmile [24]3 years ago
4 0

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

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