Answer:
The value of Δ
for the reaction = 463
![(KJ)/(mol)](https://img.qammunity.org/2021/formulas/physics/high-school/da229bukj2ttdzc47pc0czfctnoln141is.png)
Step-by-step explanation:
![q_(rxn) = - q_(sol)](https://img.qammunity.org/2021/formulas/chemistry/college/9wz0c3s5syp69l69y442wkyze2le2kdcpe.png)
![q_(rxn) = H_(rxn)](https://img.qammunity.org/2021/formulas/chemistry/college/9e1rm8ltl6pgkhce4sqldfvwvzvfszh9p3.png)
Δ
= -
------ (1)
We know that
![q_(sol) = m c ( T_(2) - T_(1) )](https://img.qammunity.org/2021/formulas/chemistry/college/4etv97hvohl8yl6mni63j2s4c5rsak5qz1.png)
(1)(100) × 4.18 × (32.8 - 25.6)
3010 J = 3.01 Kilo Joule
From equation (1)
Δ
= 3.01 Kilo Joule
No. of moles
![N = (m)/(M)](https://img.qammunity.org/2021/formulas/chemistry/college/5fazt67u9xea4r5cnp1jckzdi10887u0mb.png)
m = 0.158 gm & M = 24.31 gm Mg
No. of moles
![N = (0.158)/(24.31)](https://img.qammunity.org/2021/formulas/chemistry/college/28zpwwhuhfliffbu5uvveeo3583r5cxe38.png)
N = 0.0065
Therefore
Δ
=
![(3.01)/(0.0065)](https://img.qammunity.org/2021/formulas/chemistry/college/1z5kct5srxz7awcftfbrb9r4f836ferkwr.png)
Δ
= 463
![(KJ)/(mol)](https://img.qammunity.org/2021/formulas/physics/high-school/da229bukj2ttdzc47pc0czfctnoln141is.png)
This is the value of Δ
for the reaction.