Can someone please walk me through this?-- 25 pts!

Calculate the change of enthalpy for the reaction CH4 (g) + NH3 (g) --> HCN (g) +3H2 (g) from the following reactions:

Reaction 1: N2 (g) + 3H2 (g) --> 2NH3 (g); Change in enthalpy(∆ H): -91.8 kJ/mol

Reaction 2: C (s, graphite) + 2H2 (g) --> CH4 (g); Change in enthalpy(∆ H): -74.9 kJ/mol

Reaction 3: H2 (g) + 2C (s, graphite) + N2 (g) --> 2HCN (g); Change in enthalpy(∆ H): +270.3 kJ/mol

I have to include the following:

>The numerical answer with correct units.

>State which reactions, if any, you had to "Flip". ("Flipping" refers to modifying one of the reactions above to make a chemical be present on the other side-- like flipping an N2 that is on the reaction side. If it needs to be on the product side, you flip the reaction. Hope that makes sense.)

>State which reactions you had to multiply, if any, to get the correct amount of the compound. Also, include how much you multiplied the reaction by.

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ardni313 ardni313 · Answer 1 · 2021-01-03T22:35:44+01:00

NH3(g) + CH4(g) ⇒ HCN + 3H2 ∆ H:255.95

Given

Reaction and the enthalpy

Required

The change of enthalpy

Solution

Reaction 1

N2 (g) + 3H2 (g) --> 2NH3 ∆ H: -91.8 kJ/mol⇒reverse

2NH3 ⇒ N2 (g) + 3H2 (g) ∆ H: +91.8 kJ/mol

Reaction 2

C (s, graphite) + 2H2 (g)⇒ CH4 (g) ∆H: -74.9 kJ/mol ⇒reverse

CH4 (g) ⇒ C (s, graphite) + 2H2 (g) ∆H: +74.9 kJ/mol ⇒ x2

2CH4 (g) ⇒ 2C (s, graphite) + 4H2 (g) ∆H: +149.8 kJ/mo

Reaction 3

H2 (g) + 2C (s, graphite) + N2 (g) ⇒ 2HCN (g);∆ H: +270.3 kJ/mol

Add up all the reactions and remove the same compound from different sides :

2NH3 ⇒ N2 (g) + 3H2 (g) ∆ H: +91.8 kJ/mol

2CH4 (g) ⇒ 2C (s, graphite) + 4H2 (g) ∆H: +149.8 kJ/mol

H2 (g) + 2C (s, graphite) + N2 (g) ⇒ 2HCN (g) ∆ H: +270.3 kJ/mol

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2NH3(g) + 2CH4(g) ⇒ 2HCN + 6H2 ∆ H: 511.9 ⇒ :2

NH3(g) + CH4(g) ⇒ HCN + 3H2 ∆ H:255.95