Add the following thermochemical equations: (1) p4(s) + 6cl2(g) => 4pcl3(g) δh = –1148 kj (2) 4pcl3(g) + 4cl2(g) => 4pcl5(g) δh =–460 kj (3) overall equation = ? δhoverall = ? select the correct δh for the overall process (equation 3)

the overall reaction is

P4(s) + 10Cl2(g)   => 4PCl5(g)

and its ΔH  is

ΔH overall =  (-1608 KJ)

Explanation:

given that

P4(s) + 6Cl2(g) => 4PCl3(g)  , ΔH1 = –1148 kJ

4PCl3(g) + 4Cl2(g) => 4PCl5(g) , ΔH2 =–460 kJ

summing both equation gives

P4(s) + 6Cl2(g) +4PCl3(g) + 4Cl2(g)   => 4PCl3(g)  +  4PCl5(g)

but 4PCl3(g) is in both sides , so is cancelled out

P4(s) + 6Cl2(g)+ 4Cl2(g)   => 4PCl5(g)

summing the Cl2(g)

P4(s) + 10Cl2(g)   => 4PCl5(g) → overall reaction

(can be verified that the reaction is balanced)

since we sum both equations the ΔH of the overall reaction is also the sum of the ΔH of the individual reactions.

ΔH  overall = ΔH1 + ΔH2 =  (–1148 kJ) + (–460 kJ) = (-1608 KJ)

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

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