A) \[\frac{3\times 3}{0.5\times {{(1.5)}^{3}}}at{{m}^{-2}}\]
B) \[0.5\times {{(1.5)}^{3}}at{{m}^{2}}\]
C) \[\frac{0.5\times {{(1.5)}^{3}}}{3\times 3}at{{m}^{2}}\]
D) \[\frac{{{(1.5)}^{3}}}{0.5}at{{m}^{-2}}\]
Correct Answer: B
Solution :
\[{{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g)\] \[1-x\] \[3-3x\] \[2x\] at equilibrium Total moles, \[1-x+3-3x+2x=4-2x=3\] (given) (Since, 4 moles = 4 atm given) \[\therefore \] \[x=0.5\] \[{{K}_{p}}\] for dissociation of \[N{{H}_{3}}=\frac{P{{N}_{2}}\times {{p}^{2}}{{H}_{2}}}{{{p}^{2}}N{{H}_{3}}}\] \[=\frac{{{\left. \left( \frac{1-0.5}{3}\times 3 \right)\times \left( \frac{3-3\times 0.5}{3} \right)\times 3 \right]}^{3}}}{{{\left[ \frac{2\times 0.5\times 3}{3} \right]}^{2}}}\] \[=0.5\times {{(1.5)}^{2}}\,at{{m}^{2}}\]
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