JEE MAIN - Chemistry (2019 - 11th January Morning Slot - No. 2)
Consider the reaction
N2(g) + 3H2(g) $$\rightleftharpoons$$ 2NH3(g)
The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that PNH3 << Ptotal at equilibrium)
N2(g) + 3H2(g) $$\rightleftharpoons$$ 2NH3(g)
The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that PNH3 << Ptotal at equilibrium)
$${{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 4}$$
$${{K_P^{{1 \over 2}}{P^2}} \over 4}$$
$${{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 16}$$
$${{K_P^{{1 \over 2}}{P^2}} \over 16}$$
Explanation
N2(g) + 3H2(g) $$\rightleftharpoons$$ 2NH3(g) ; Keq = Kp
Write this equation reverse way,
2NH3(g) $$\rightleftharpoons$$ N2(g) + 3H2(g) ; Keq = $${1 \over {{K_p}}}$$
At equillibrium
PTotal = PNH3 + PN2 + PH2
= PNH3 + p + 3p
(As PNH3 << Ptotal so we can ignore PNH3)
$$ \therefore $$ PTotal = 4p
$$ \Rightarrow $$ p = $${{{{P_{total}}} \over 4}}$$
Formula of
Keq = $${{{p_{{N_2}}} \times {{\left( {{p_{{H_2}}}} \right)}^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$$ = $${1 \over {{K_p}}}$$
$$ \Rightarrow $$ $${1 \over {{K_p}}}$$ = $${{p \times 27{p^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$$
$$ \Rightarrow $$ $${{{\left( {{p_{N{H_3}}}} \right)}^2}}$$ = Kp $$ \times $$ 27 $$ \times $$ $${{{\left( {{{{P_{total}}} \over 4}} \right)}^4}}$$
$$ \Rightarrow $$ PNH3 = $$\sqrt {{K_p}} \times {\left( {27} \right)^{{1 \over 2}}} \times {\left( {{{{P_{Total}}} \over 4}} \right)^{{4 \over 2}}}$$
= $${{{3^{{3 \over 2}}}K_p^{{1 \over 2}}P_{Total}^2} \over {16}}$$
Write this equation reverse way,
2NH3(g) $$\rightleftharpoons$$ N2(g) + 3H2(g) ; Keq = $${1 \over {{K_p}}}$$
| 2NH3(g) | ⇌ | N2(g) | + | 3H2(g) | |
|---|---|---|---|---|---|
| At t = 0 | Po | 0 | 0 | ||
| At t = teq | PNH3 | p | 3p |
At equillibrium
PTotal = PNH3 + PN2 + PH2
= PNH3 + p + 3p
(As PNH3 << Ptotal so we can ignore PNH3)
$$ \therefore $$ PTotal = 4p
$$ \Rightarrow $$ p = $${{{{P_{total}}} \over 4}}$$
Formula of
Keq = $${{{p_{{N_2}}} \times {{\left( {{p_{{H_2}}}} \right)}^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$$ = $${1 \over {{K_p}}}$$
$$ \Rightarrow $$ $${1 \over {{K_p}}}$$ = $${{p \times 27{p^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$$
$$ \Rightarrow $$ $${{{\left( {{p_{N{H_3}}}} \right)}^2}}$$ = Kp $$ \times $$ 27 $$ \times $$ $${{{\left( {{{{P_{total}}} \over 4}} \right)}^4}}$$
$$ \Rightarrow $$ PNH3 = $$\sqrt {{K_p}} \times {\left( {27} \right)^{{1 \over 2}}} \times {\left( {{{{P_{Total}}} \over 4}} \right)^{{4 \over 2}}}$$
= $${{{3^{{3 \over 2}}}K_p^{{1 \over 2}}P_{Total}^2} \over {16}}$$
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