The reaction 2 NO(g) + Cl_2(I) rightarrow 2 NOCl(g) has DeltaHdegree = -77.4 kJ

Question

The reaction 2 NO(g) + Cl_2(I) rightarrow 2 NOCl(g) has DeltaHdegree = -77.4 kJ What are the DeltaHdegree for the following: a) NO(g) + 1/2 Cl_2(g) rightarrow NOCl(g) DeltaHdegree = b) 6 NOCl(g) rightarrow 6NO(g) + 3Cl_2(g) DeltaHdegree = 16. Given the following thermochemical equations: 2 Cu + S rightarrow Cu_2S DeltaHdegree = -79.5 kJ S + O_2 rightarrow SO_2 DeltaHdegree = -297 kJ Cu_2S + 2 O_2 rightarrow 2 CuO + SO_2 DeltaHdegree = - 527.5 kJ Calculate the value of DeltaHdegree in kJ for the formation of CuO from the elements. Given the following thermochemical equations: 3 Mg + 2 NH_3 rightarrow Mg_3N_2 + 3 H_2 DeltaHdegree = -371 kJ 1/2 N_2 + 3/2 H_2 rightarrow NH_3 DeltaHdegree = -46 kJ Calculate DeltaHdegree for 3 Mg + N_2 rightarrow Mg_3N_2

Explanation / Answer

1)

a.

we need the half reaction, so

Hnew = H/2 = -77.4/2 = 38.7 kJ/mol

b.

this is:

6 times larger, and inverted so overall effect:

Hrxn = (-1)(6)*H = -6*(-77.4) = 464.4 kJ/mol

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