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Consider these reactions where M represents a generic metal: 2M(s) + 6HCl(aq) ---->2MCl3(aq) + 3H2(g) Delta...

Consider these reactions where M represents a generic metal:

2M(s) + 6HCl(aq) ---->2MCl3(aq) + 3H2(g) Delta H1 = -569kj

HCl(g) ----> HCl(aq)   DeltaH2 = -74.8

H2(g) + Cl2(g) ----> 2HCl(g) Delta H3 = -1845kj

MCl3(s) ----> MCl3(aq)   Delta H4 = -405kj

Use the above information to determine the enthalpy of the following reaction:

2M(s) + 3Cl2(g) ----> 2MCl3(s)   What is Delta H for this reaction?

Solutions

Expert Solution

notes:

use hess laws for enthalpy changes

note that we need

2M(s) in the left, so eqn 1 remains the same

we need to cancel HCl(aq) so, we need to multiply eqn 2 by (6)

then

2M(s) + 6HCl(aq) ---->2MCl3(aq) + 3H2(g) Delta H1 = -569

6HCl(g) ----> 6HCl(aq)   DeltaH2 = -74.8*6 = -448.8

Add both equations

2M(s) + 6HCl(aq)+6HCl(g) ---->2MCl3(aq) + 3H2(g) + 6HCl(aq) dHRxn = -569+-448.8 = -1017.8

cancel common terms

2M(s) +6HCl(g) ---->2MCl3(aq) + 3H2(g) Hrxn = -1017.8

Now, we need to ancel HCl and H2

Eqn 3 must be mutiplied by 3

2M(s) +6HCl(g) ---->2MCl3(aq) + 3H2(g) Hrxn = -1017.8

3H2(g) + 3Cl2(g) ----> 6HCl(g) Delta H3 = -1845*3 = -5535

add both equations

2M(s) +6HCl(g) +3H2(g) + 3Cl2(g) ----->2MCl3(aq) + 3H2(g) +6HCl(g) Hrxn = -1017.8+ -5535 = -6552.8

cancel common terms

2M(s) + 3Cl2(g) ----->2MCl3(aq) Hrxn = = -6552.8

Now, we need MCl3 in solid state not aquous so...

from equation 4, invert it and mutiply by 2

2M(s) + 3Cl2(g) ----->2MCl3(aq) Hrxn = = -6552.8

2MCl3(aq) ---->2MCl3(s) Delta H4 = +405*2 = 810

Add both

2M(s) + 3Cl2(g)+2MCl3(aq)----->2MCl3(aq) +2MCl3(s) Hrxn = = -6552.8+810

cancel common terms

2M(s) + 3Cl2(g)----->2MCl3(s) Hrxn = = -5742.8 kJ

which is what we wanted

queen_honey_blossom answered 2 years ago
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