Class 11th

K_p = 47.9      T = 288 K

K_C =?                    R = 0.083 L bar/K mol

$N_2{O}_4\left(g\right)?2N{O}_2\left(g\right)$           

Using

$K_p=K_C{\left(RT\right)}^{\Delta ng}$              

Here ;

$\Delta ng=1$                

49.7 = K_C (0.083 * 288)¹

K_C = 2

More enthalpy of combustion of graphite is -2.48 * 10² kJ/mol, that means 2.48 * 10² kJ heat generates when combustion of 1 mol graphite takes place.

So, heat generated on combustion of 1 g graphite

$=\frac{2.48*10^2}{12}kJ$

= 20.67 kJ

Ans. is 21 (the nearest integer)

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