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](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DH_2%28g%29%20%2B%5Cfrac%7B1%7D%7B2%7DBr_2%28l%29%5Crightarrow%20HBr%28g%29%20%2C%5CDelta%20H_%7Bf%7D%5Eo%3D%20-36.29%20kJ)
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
gas and 1 mole of
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](https://tex.z-dn.net/?f=H_2%28g%29%20%2B%20Br_2%28l%29%5Crightarrow%202HBr%28g%29%20%2C%5CDelta%20H_%7Bf%7D%5Eo%3D%20-72.58kJ)
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](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DH_2%28g%29%20%2B%5Cfrac%7B1%7D%7B2%7DBr_2%28l%29%5Crightarrow%20HBr%28g%29%20%2C%5CDelta%20H_%7Bf%7D%5Eo%3D%20-36.29%20kJ)
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](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DH_2%28g%29%20%2B%5Cfrac%7B1%7D%7B2%7DBr_2%28l%29%5Crightarrow%20HBr%28g%29%20%2C%5CDelta%20H_%7Bf%7D%5Eo%3D%20-36.29%20kJ)
Kc = concentrations of product / concentrations of reactant
Kc = [Br₂] [Cl₂]₃ / [BrCl₃]₂
What is the equilibrium constant?
The relationship between a reaction's products and reactants with regard to a certain unit is expressed by the equilibrium constant(K) This article introduces the mathematics needed to determine the partial pressure equilibrium constant as well as how to formulate expressions for equilibrium constants. By allowing a single reaction to reach equilibrium and then measuring the concentrations of each chemical participating in that reaction, one can determine the numerical value of an equilibrium constant. it is the ratio of product concentrations to reactant concentrations. The equilibrium constant for a given reaction is unaffected by the initial concentrations because the concentrations are measured at equilibrium.
To learn more about the equilibrium constant, visit:
brainly.com/question/19340344
#SPJ4
Nitrogen is more reactive than oxygen and oxygen than chlorine
Answer: first option is not a termination
∙CH3 + Cl2 → CH3Cl + Cl∙
Explanation:
Since a radical is formed as part of the product it means it's a propagation step and not a termination step, at termination no free radical exist as product
Answer:
Iron is oxidized while chlorine is reduced.
Explanation:
The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reaction:
2FeCl₂ + Cl₂ → 2FeCl₃
in this reaction the oxidation state of iron is increased from +2 to +3. That's why iron get oxidized and it is reducing agent because it reduced the chlorine. The chlorine is reduced from -2 to -3 and it is oxidizing agent because it oxidized the iron.
2Fe⁺²Cl₂⁻²
2Fe⁺³Cl₃⁻³
The iron atom gives it three electrons to three atoms of chlorine and gain positive charge while chlorine atom accept the electron and form anion.