Question

Calculate the value of kp for the reaction $$2n2(g)+o2(g)2n2o(g) at 298.15 k and 1773 k. Thermodynamic data for n2o(g) are: hf = 82.05 kj/mol; s = 219.9 j/mol k; gf = 104.2 kj/mol

Answer 1

Calculate the value of kp for the reaction $$2n2(g)+o2(g)2n2o(g) at 298.15 k and 1773 k. Thermodynamic data for n2o(g) are: hf = 82.05 kj/mol; s = 219.9 j/mol k; gf = 104.2 kj/mol

Answer 2

Explanation:

Considering the equation

$2N_{2}O$ --> $2N_{2} + O_{2}$

For the above reaction find the value of Gr

ΔGr = (ΔGf $N_{2}$ x 2 + 2Gf$O_{2}$) - 2ΔGf $NO_{2}$

       = 0 + 0 - 2 x 104.2 Kj

ΔGr = -208.4 Kj/mol

ΔGr = -2.303RT logkp

putting the values in the equation

-208.4 x $10^{3}$ = -2.303 x 8.314 x 1773 logkp

rearranging the terms

log kp = $\frac{208.4 * 1000}{2.303 * 8.314 * 1773}$

log kp = 6.1388

taking antilog

kp = 1.37 x $10^{6}$