The reaction between hydrogen gas and chlorine gas produces hydrogen chloride according to the following equation: H2(g) + Cl2(g) → 2HCl(g). Using the following bond enthalpies(D) calculate the enthalpy of reaction. D(H-H) =436 kJ/mol; D(Cl-Cl) = 243 kJ/mol D(H-Cl) = 432 kJ/mol

The enthalpy of reaction is -185 kJ

Explanation:

To get the reaction:

 H₂(g) + Cl₂(g) → 2 HCl(g)

you must follow the following steps:

1)  Reactive molecules must break their bonds to obtain their atoms.

H₂(g) → 2 H(g)

Cl₂(g) → 2 Cl(g)

Bond energy (or enthalpy) is the energy required to break one mole of bonds of a gaseous substance. In the case of diatomic molecules with a single bond, it corresponds to the energy necessary to dissociate 1 mole of said substance in the atoms that form it.

Whenever you want to break links you must supply energy, so the link enthalpy will have positive values; while when a mole of bonds is formed energy is released and the bond enthalpy of this process will be negative.

In this case you will then have:

H₂(g) → 2 H(g)           ΔH=436 kJ/mol

Cl₂(g) → 2 Cl(g)         ΔH=243 kJ/mol

So the total energy needed to break all the bonds is:

ΔH=1 mol*436 kJ/mol +1 mol* 243 kJ/mol= 679 kJ

2) The atoms that were obtained in the break of the bonds must be combined to obtain the product.

2 H (g) + 2 Cl (g) → 2 HCl (g)

Being the single bond energy for one mole of 431 kJ H-Cl bonds and considering that two moles of H-Cl bonds are formed, the ΔH is:

ΔH = -2 moles* (432 kJ/mol) = -864 kJ

As mentioned, when a mole of bonds is formed energy is released, the bond enthalpy of this process will be negative.  So the formation of HCl is negative.

Hess's law states that the energy change in an overall chemical reaction is equal to the sum of the energy changes in the individual reactions comprising it. So:

ΔHtotal= -864 kJ + 679 kJ

ΔHtotal= -185 kJ

The enthalpy of reaction is -185 kJ

what are the statements

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2