Question

For a reaction, 2X(s) +2Y(s) -> 2C(l)+D(g)The qp at 27°C is-28 Kcal mol-. The value of qv is:

Answer 1

We know that,

$q_{p} = H$

$q_{v} = E$

H = E + nRT

for given reaction n = product (gaseous species) - reactant (gaseous species)

Therefore,

n =  1-0=1

Substituting given values in above relation

E = H - nRT = [ 28000 - (1×1.987×300) ]

  = 28000 - 596.1 = 27403.9 cal/mol

[tex]q_{v} or E   = 27.404 Kcal/mol

Answer 2

∆U = ∆H - ∆nRT

∆n = 1 - 0 -1

∆n = 1

Then value of R = 2 cal k mol

T = 300 k

therefore,

∆H = ∆v + ∆ n RT

-28 = ∆U + 1 × 600

-28 × 10 power 3 = ∆U + 600

-28000 = ∆U + 600

the final answer is.. -28.6 kcal

therefore the ∆V = -28.6 kcal