JEE MAIN - Chemistry (2005 - No. 7)
For the reaction
2NO2 (g) $$\leftrightharpoons$$ 2NO (g) + O2 (g), (Kc = 1.8 $$\times$$ 10-6 at 184oC) (R = 0.0831 kJ/(mol. K))
When Kp and Kc are compared at 184oC , it is found that :
When Kp and Kc are compared at 184oC , it is found that :
Kp is greater than Kc
Kp is less than Kc
Kp = Kc
Whether Kp is greater than, less than or equal to Kc depends upon the total gas
pressure
Explanation
For the reaction : -
$$2N{O_2}\left( g \right)\rightleftharpoons2NO\left( g \right) + {O_2}\left( g \right)$$
Given $${K_c} = 1.8 \times {10^{ - 6}}\,\,$$ at $$\,\,{184^ \circ }C$$
$$R=0.0831$$ $$\,\,kJ/mol.k$$
$${K_p} = 1.8 \times {10^{ - 6}} \times 0.0831 \times 457$$
$$ = 6.836 \times {10^{ - 6}}$$
$$\left[ {} \right.$$ as $$\,\,\,\,\,{184^ \circ }C = \left( {273 + 184} \right) = 457\,k,\,\,$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. {\Delta n = \left( {2 + 1, - 1} \right) = 1} \right]$$
Hence it is clear that $${K_p} > {K_c}$$
$$2N{O_2}\left( g \right)\rightleftharpoons2NO\left( g \right) + {O_2}\left( g \right)$$
Given $${K_c} = 1.8 \times {10^{ - 6}}\,\,$$ at $$\,\,{184^ \circ }C$$
$$R=0.0831$$ $$\,\,kJ/mol.k$$
$${K_p} = 1.8 \times {10^{ - 6}} \times 0.0831 \times 457$$
$$ = 6.836 \times {10^{ - 6}}$$
$$\left[ {} \right.$$ as $$\,\,\,\,\,{184^ \circ }C = \left( {273 + 184} \right) = 457\,k,\,\,$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. {\Delta n = \left( {2 + 1, - 1} \right) = 1} \right]$$
Hence it is clear that $${K_p} > {K_c}$$
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