Question

0 Standard Enthalpies of Formation at 25°C Substance AH; (kJ/mol) B(s) B2O3(s) – 1273.5 BF3(g) -1136.0 C(s, graphite) C(s, diamond) 1.88 C2H2(g) 227.4 C6H6(1) 49.1 HF(9) –273.3 F(g) 79.38 472.7 0 91.3 249.2 N(g) N2(g) NO(g) O(g) O2(g) 03 (9) H(9) H2(g) 142.7 218.0 O

Part A:

Determine ΔH∘f for NO(g)

Part B:

Determine ΔH∘f for BF3(g)

Part C:

Determine ΔH∘f for C6H6(l)

Part D:

Determine ΔH∘f for HF(g)

Answer 1

A)

Standard state of N is N2(g)

Standard state of O is O2(g)

So, the formation reaction is:

1/2 N2(g) + 1/2 O2(g) -> NO(g)

From table:

ΔHof(NO(g)) = 91.3 KJ/mol

Answer: 91.3 KJ/mol

B)

Standard state of B is B(s)

Standard state of F is F2(g)

So, the formation reaction is:

B(s) + 3/2 F2(g) -> BF3(g)

From table:

ΔHof(BF3(g)) = -1136.0 KJ/mol

Answer: -1136.0 KJ/mol

C)

Standard state of C is C(s,graphite)

Standard state of H is H2(g)

So, the formation reaction is:

6 C(s,graphite) + 3 H2(g) -> C6H6(l)

From table:

ΔHof(C6H6(l)) = 49.1 KJ/mol

Answer: 49.1 KJ/mol

D)

Standard state of H is H2(g)

Standard state of F is F2(g)

So, the formation reaction is:

1/2 H2(g) + 1/2 F2(g) -> HF(g)

From table:

ΔHof(HF(g)) = -273.3 KJ/mol

Answer: -273.3 KJ/mol