Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

For the reaction

$2NO_{2(g)}\rightleftharpoons 2NO_{(g)}+O_{2(g)}$

$\left ( K_{c}= 1.8\times 10^{-6}\: at\: 184^{\circ}C\right )$

$\left ( R= 0.0831\: kJ/\left ( mol.K \right ) \right )$ When $K_{p}and K_{c}$ are compared at $184^{\circ}C$ it is found that

Option 1)
$K_{p}\: is\: greater\: than\: K_{c}$

Option 2)
$K_{p}\: is\: less\: than\: K_{c}$

Option 3)
$K_{p}\: = \: \: K_{c}$

Option 4)
$Whether \: K_{p}\: is\: greater\: than\: ;less \: than \: or\: equal\: K_{c}$ $depends\: upon\: the\: total\: gas\: pressure\:$

Answers (1)

best_answer

As we learnt in

Relation between Kp and Kc -

$K_{p}=K_{c}(RT)^{\bigtriangleup n}$

$\therefore K_{p}> K_{c}$ while calculating the value of K_p , pressure should be expressed in bar.

$1\:bar=10^{5}pa$

- wherein

$\bigtriangleup n$ = (number of moles of gaseous products) - (number of moles gaseous reaction)

$2NO_{2}\left _{\left ( s \right )} \: \rightleftharpoons 2NO \: _{\left ( g \right )}+ O_{2} \: _{\left ( g \right )}$

$K_{c}+1.8\times 10^{-6} \: at \: 457K$

$K_{p}=K_{c}(RT)^{\Delta n}$ where $\Delta n=3-2=1$

$K_{p}=K_{c}(0.0831\times 457)^{1}\; \; \; \; \;$

$K_{p}=1.8\times 10^{-6}\times (0.0831\times 457)^{1}$

$K_{p}=6.8\times 10^{-5}$

$\therefore K_{p}> K_{c}$

Option 1)

$K_{p}\: is\: greater\: than\: K_{c}$

This option is correct

Option 2)

$K_{p}\: is\: less\: than\: K_{c}$

This option is incorrect

Option 3)

$K_{p}\: = \: \: K_{c}$

This option is incorrect

Option 4)

$Whether \: K_{p}\: is\: greater\: than\: ;less \: than \: or\: equal\: K_{c}$ $depends\: upon\: the\: total\: gas\: pressure\:$

This option is incorrect

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question