JEE MAIN - Chemistry (2020 - 6th September Evening Slot - No. 14)
The value of KC is 64 at 800 K for the reaction
N2(g) + 3H2(g) ⇌ 2NH3(g)
The value of KC for the following reaction is :
NH3(g) ⇌ $${1 \over 2}$$N2(g) + $${3 \over 2}$$H2(g)
N2(g) + 3H2(g) ⇌ 2NH3(g)
The value of KC for the following reaction is :
NH3(g) ⇌ $${1 \over 2}$$N2(g) + $${3 \over 2}$$H2(g)
8
$${1 \over 8}$$
$${1 \over 4}$$
$${1 \over {64}}$$
Explanation
N2(g) + 3H2(g) ⇌ 2NH3(g) ; KC
2NH3(g) ⇌ N2(g) + 3H2(g) ; $${1 \over {{K_C}}}$$
Multiplying by $${1 \over 2}$$, reaction becomes
NH3(g) ⇌ $${1 \over 2}$$N2(g) + $${3 \over 2}$$H2(g) ;
$$ \therefore $$ New KC = $${\left( {{1 \over {{K_C}}}} \right)^{{1 \over 2}}}$$ = $${\left( {{1 \over {64}}} \right)^{{1 \over 2}}}$$ = $${1 \over 8}$$
2NH3(g) ⇌ N2(g) + 3H2(g) ; $${1 \over {{K_C}}}$$
Multiplying by $${1 \over 2}$$, reaction becomes
NH3(g) ⇌ $${1 \over 2}$$N2(g) + $${3 \over 2}$$H2(g) ;
$$ \therefore $$ New KC = $${\left( {{1 \over {{K_C}}}} \right)^{{1 \over 2}}}$$ = $${\left( {{1 \over {64}}} \right)^{{1 \over 2}}}$$ = $${1 \over 8}$$
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