question_answer 1)

13.8 g of \[{{N}_{2}}{{O}_{4}}\](g) was placed in a 1L reaction vessel at 400 K and allowed to attain equilibrium: \[{{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)\] The total pressure at equilibrium was found to be 9.15 bar. Calculate \[{{K}_{c}},{{K}_{p}}\] and partial pressure at equilibrium.

Answer:

Number of moles of\[{{N}_{2}}{{O}_{4}}\], \[n=\frac{mass}{molecular\,mass}\] \[=\frac{13.8}{92}=0.15\] \[PV=nRT\] \[P\times 1=0.15\times 0.083\times 400\] \[P=4.98bar\] The reaction is: \[\underset{{{t}_{eq}}}{\overset{{}}{\mathop{{{t}_{0}}}}}\,\underset{\underset{(4.98-x)}{\mathop{4.98}}\,}{\mathop{{{N}_{2}}{{O}_{4}}(g)}}\,\rightleftharpoons \underset{\underset{2x}{\mathop{0}}\,}{\mathop{2N{{O}_{2}}(g)}}\,\] Total pressure = \[{{p}_{{{N}_{2}}{{O}_{4}}}}+{{p}_{N{{O}_{2}}}}\] \[=4.98-x+2x=(4.98+x)\] \[9.15=4.98+x\] \[x=4.17bar\] \[{{p}_{N{{O}_{2}}}}=2x=2\times 4.17=8.34bar\] \[{{p}_{{{N}_{2}}{{O}_{4}}}}=4.98-x=4.98-4.17=0.81bar\] \[{{K}_{p}}\frac{{{({{p}_{N{{O}_{2}}}})}^{2}}}{({{p}_{{{N}_{2}}{{O}_{4}}}})}\] \[=\frac{{{(8.34)}^{2}}}{0.81}=85.87bar\] We know, \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\] \[{{K}_{c}}=\frac{{{K}_{p}}}{{{(RT)}^{\Delta {{n}_{g}}}}}=\frac{85.87}{{{(0.083\times 400)}^{1}}}\] =2.586= 2.6 M