Question

Given `PCl_5(g) rarr PCl_3(g) +Cl_2(g)` `DeltaH^@=+67*kJ*mol^-1`. What energy must be input to decompose a mass of `282*g` of `PCl_5`?

What energy must be input to decompose a mass of `282*g` of `PCl_5`?

Answer 1

We first write out the equation......

Explanation:

`PCl_5(g) +Delta rarr PCl_3(g) + Cl_2(g)`....or

`PCl_5(g) rarr PCl_3(g) + Cl_2(g);` `DeltaH_"rxn"^@=+67*kJ*mol^-1`

The positive charge of `DeltaH^@` denotes that this is an endothermic reaction; and this is reasonable in that we are BREAKING bonds, which typically require the input of energy.

Now our data always refers to moles of reaction as written.....so we KNOW that the decomposition of `1*mol` `PCl_5(g)`, `(208.2*g)` requires the INPUT of `67*kJ` of energy to produce `1*mol` of `PCl_3(g)`, `137.3*g`, and `1*mol` of `Cl_2(g)`, `70.9*g`.

We start with a molar quantity with respect to `PCl_5` of....

`(282*g)/(208.24*g*mol^-1)=1.35*mol`...

And we have been given `DeltaH_"rxn"^@=+67*kJ*mol^-1`. And so we take the product...

`+67*kJ*mol^-1xx1.35*mol-=90.7*kJ`...and the positive sign of `DeltaH_"rxn"^@` denotes that energy must PUT IN to the reaction for it to proceed.

Do you follow? If there is a point of confusion, I and others will be happy to try to clarify.