Question

What is the answer to this Chem question

Answer 1

The enthalpy change (ΔH) for the reaction 2Sb(s) +
`5Cl_2`(g) →
`2SbCl_5`(g) can be calculated using Hess's Law and is found to be +468 kJ, which corresponds to option b.

To find ΔH for the reaction of 2Sb(s) +
`5Cl_2`(g) →
`2SbCl_5`(g), we need to use Hess's Law and combine the given steps to match the overall reaction. We have two given reactions, the first one being the formation of
`SbCl_3` from Sb and
`Cl_2`, and the second being the formation of
`SbCl_5` from
`SbCl_3` and
`Cl_2`:

• 2Sb(s) +
  `3Cl_2`(g) →
  `2SbCl_3`(g) ΔΔH = +628 kJ
• 
  `SbCl_3`(g) +
  `Cl_2`(g) →
  `2SbCl_5`(g) ΔΔH = -80 kJ

The overall reaction can be found by adding the first reaction as it is and the second reaction multiplied by 2 (to account for 2 moles of Sb), giving us:

• 2Sb(s) +
  `3Cl_2`(g) →
  `2SbCl_3`(g)
• 
  `2SbCl_3`(g) +
  `2Cl_2`(g) →
  `2SbCl_5`(g)

Add these reactions:

• 2Sb(s) +
  `5Cl_2`(g) → 2SbCl5(g)

Then, add the enthalpy changes: +628 kJ + 2(-80 kJ) = +628 kJ - 160 kJ = +468 kJ.

Hence, the ΔΔH for the reaction is +468 kJ.