Calculate delta H for the reaction CH4 (g) + NH3 (g) --> HCN (g) + 3 H2 (g), giv

Q: Calculate delta H for the reaction CH4 (g) + NH3 (g) --> HCN (g) + 3 H2 (g), giv

Solution 1 CH4 (g) + NH3 (g) --> HCN (g) + 3 H2 (g) delta H = -Reaction 1 * 1/2 + (-Reaction 2) + (1/2 * reaction 3 ) = +91.8/2 + (74.9) + (270.3/2) = 255.95 KJ 2 Al (s) + 3 Cl2 (g) --> 2 AlCl3 (s) delta H = Reaction 1 + 6* reaction 2 + 3* reaction 3 + (-Reaction 4 * 2 ) = (-1049) + (6*-74.8) + (3*-1845) + (2*323) = -6386.8 KJ

ID: 905194 • Letter: C

Question

Calculate delta H for the reaction CH4 (g) + NH3 (g) --> HCN (g) + 3 H2 (g), given:

N2 (g) + 3 H2 (g) --> 2 NH3 (g) delta H = -91.8 kJ

C (s) + 2 H2 (g) --> CH4 (g) dleta H = -74.9 kJ

H2 (g) + 2 C (s) + N2 (g) --> 2 HCN (g) delta H = +270.3 kJ

Calculate delta H for the reaction 2 Al (s) + 3 Cl2 (g) --> 2 AlCl3 (s) from the ata.

2 Al (s) + 6 HCl (aq) --> 2 AlCl3 (aq) + 3 H2 (g) delta H = -1049. kJ

HCl (g) --> HCl (aq) deltaH = -74.8 kJ

H2 (g) + Cl2 (g) --> 2 HCl (g) delta H = -1845. kJ

AlCl3 (s) --> AlCl3 (aq) delta H = -323. kJ

Please provide an explanation for the breakdown of the necessary steps

Explanation / Answer

Solution 1

CH4 (g) + NH3 (g) --> HCN (g) + 3 H2 (g)

delta H = -Reaction 1 * 1/2 + (-Reaction 2) + (1/2 * reaction 3 )

= +91.8/2 + (74.9) + (270.3/2)

= 255.95 KJ

2 Al (s) + 3 Cl2 (g) --> 2 AlCl3 (s)

delta H = Reaction 1 + 6* reaction 2 + 3* reaction 3 + (-Reaction 4 * 2 )

= (-1049) + (6*-74.8) + (3*-1845) + (2*323)

= -6386.8 KJ

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Calculate delta H for the reaction CH4 (g) + NH3 (g) --> HCN (g) + 3 H2 (g), giv

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