Question

For a gaseous reaction, a(g)+3b(g)3c(g)+3d(g), u is 17kcal at 27oc. Assuming r=2 cal k1 mol1, the value of h for the above reaction is:

Answer 1

H = U + nRT

= 17 + 2(2)(300)×10-³

= 17 + 1.2

= 18.2 kcal.

Answer 2

The value of $\Delta H$ for the above reaction is 18.2 kcal

Explanation:

For the given reaction:

$a(g)+3b(g)\rightarrow 3c(g)+3d(g)$

$\Delta H=\Delta U+\Delta n_g\times RT$

where,

$\Delta H$ =  enthalpy of the reaction= ?

$\Delta U$= internal energy of the reaction = 17 kcal = 17000 cal (1kcal=1000cal)

$\Delta n_g$ = change in the moles of the reaction = Moles of product - Moles of reactant = 6 - 4 = 2 mole

R = gas constant = 2 cal/moleK

T = temperature = $27^0C=(27+273)K=300K$

Putting in the values we get:

$\Delta H=17000+2\times 2\times 300$

$\Delta H=18200cal=18.2kcal$

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