JEE MAIN - Chemistry (2011 - No. 5)
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on
the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :
3 atm
0.3 atm
0.18 atm
1.8 atm
Explanation
$$C{O_2} + {C_{\left( {grapnite} \right)}}\,\rightleftharpoons\,2CO$$
$${P_{initial}}\,\,0.5atm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$$
$${P_{final}}\,\,\,\left( {0.5 - x} \right)atm\,\,\,\,\,\,\,2x\,atm$$
Total $$P$$ at equilibrium
$$ = 0.5 - x + 2x = 0.5 + x\,atm$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.8 = 0.5 + x$$
$$\therefore$$ $$\,\,\,x = 0.8 - 0.5 = 0.3\,atm$$
Now $$\,\,\,{k_p} = {\left( {{P_{CO}}} \right)^2}/{P_{C{O_2}}}$$
$$ = {{{{\left( {2 \times 0.3} \right)}^2}} \over {\left( {0.5 - 0.3} \right)}} = {{{{\left( {0.6} \right)}^2}} \over {\left( {0.2} \right)}}$$
$$ = 1.8\,atm$$
$${P_{initial}}\,\,0.5atm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$$
$${P_{final}}\,\,\,\left( {0.5 - x} \right)atm\,\,\,\,\,\,\,2x\,atm$$
Total $$P$$ at equilibrium
$$ = 0.5 - x + 2x = 0.5 + x\,atm$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.8 = 0.5 + x$$
$$\therefore$$ $$\,\,\,x = 0.8 - 0.5 = 0.3\,atm$$
Now $$\,\,\,{k_p} = {\left( {{P_{CO}}} \right)^2}/{P_{C{O_2}}}$$
$$ = {{{{\left( {2 \times 0.3} \right)}^2}} \over {\left( {0.5 - 0.3} \right)}} = {{{{\left( {0.6} \right)}^2}} \over {\left( {0.2} \right)}}$$
$$ = 1.8\,atm$$
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