PLEASE HELP!!!!!!!

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°

miskamm [114]

Answer: The $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)$ $\Delta H_1=157kJ$ ( × 4)

(2) $P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)$ $\Delta H_2=-1207kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

6 0

3 years ago

A 29.7 g sample of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to

Softa [21]

Answer: The mass of iron in the ore is 10.9 g

Explanation:

We are given:

Mass of iron (III) oxide = 15.6 g

We know that:

Molar mass of Iron (III) oxide = 159.69 g/mol

Molar mass of iron atom = 55.85 g/mol

As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.

To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:

In 159.69 g of iron (III) oxide, mass of iron present is $(2\times 55.85)=111.7g$