Question

Fir a gaseous reaction a+3b=3c+3d. ∆e is 17 k cal at 27°c assuming r=2 the value of ∆h for the above reaction

Answer 1

Answer : The value of ΔH for above reaction is 18200 cal.

Solution : Given,

ΔE = 17 K cal = (17 × 1000) cal = 17000 cal                  (1 K cal = 1000 cal)

Temperature = $27^{0}C$ = 27 + 273 = 300 K           ($1^{0}C$ = 273 K)

R = 2 cal/mole K

Formula used :  

$\Delta H=\Delta E+\Delta n_g\times R\times T$           .............(1)

where,

$\Delta H$ = enthalpy of reaction

$\Delta E$ = internal energy of reaction

$\Delta n_g$ = number of moles

R = Gas constant

T = temperature

The given gaseous reaction is,

a + 3b = 3c + 3d

First we have to calculate the number of moles by the given reaction.

$\Delta n_g$ = moles of product - moles of reactant = 6 - 4 = 2 moles

Now put all the given values in above formula (1), we get

$\Delta H=17000cal+2 moles\times 2cal/moleK\times 300K$

$\Delta H$ = 18200 cal

Answer 2

A(g)+3B(g)----->3C(g)+4D(g)
Formula used:
∆H=∆E+∆ngRT. ∆ng=6-4=2
=17 + 2×2×10^-3×300
=18.2 kcal