Explain why the partial pressure $p$ of a gas is proportional to its concentration $c$.
(N/A) For reactions involving gases,it is often convenient to express the equilibrium constant in terms of partial pressure.
The ideal gas equation is given by $pV = nRT$ (Eq. $i$).
Rearranging for pressure,we get $p = (n/V)RT$ (Eq. $ii$).
Since concentration $c$ is defined as $n/V$ (in $mol \ L^{-1}$ or $mol \ dm^{-3}$),we can substitute this into the equation:
$p = cRT$ (Eq. $iii$).
Here,$R$ is the universal gas constant,$T$ is the temperature in Kelvin,and $c$ is the molar concentration of the gas.
At a constant temperature $(T)$,the product $RT$ is constant.
Therefore,the partial pressure $p$ of the gas is directly proportional to its concentration $c$,expressed as $p \propto c$ or $p \propto [\text{gas}]$.
The ideal gas equation is given by $pV = nRT$ (Eq. $i$).
Rearranging for pressure,we get $p = (n/V)RT$ (Eq. $ii$).
Since concentration $c$ is defined as $n/V$ (in $mol \ L^{-1}$ or $mol \ dm^{-3}$),we can substitute this into the equation:
$p = cRT$ (Eq. $iii$).
Here,$R$ is the universal gas constant,$T$ is the temperature in Kelvin,and $c$ is the molar concentration of the gas.
At a constant temperature $(T)$,the product $RT$ is constant.
Therefore,the partial pressure $p$ of the gas is directly proportional to its concentration $c$,expressed as $p \propto c$ or $p \propto [\text{gas}]$.
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For the given equilibrium reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,choose the correct equation to calculate $K_p$.
The gas phase reaction $2A_{(g)} \rightleftharpoons A_{2(g)}$ at $400 \ K$ has $\Delta G^{\circ} = +25.2 \ kJ \ mol^{-1}$. The equilibrium constant $K_{C}$ for this reaction is $...... \times 10^{-2}$. (Round off to the nearest integer) $[$Use: $R = 8.3 \ J \ mol^{-1} \ K^{-1}$,$\ln 10 = 2.3$,$\log_{10} 2 = 0.30$,$1 \ atm = 1 \ bar]$ $[$antilog $(-3.3) = 5.01 \times 10^{-4}]$
Given below are certain reactions. Identify the reaction for which $K_p = K_c$.
MediumNEET 2026
View SolutionThe values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below: $A(g) \rightleftharpoons B(g) + C(g)$.
$1/T \text{ (K}^{-1})$ $\log_{10} K_p$ $0.05$ $3.5$ $0.06$ $2.5$ $0.07$ $1.5$
The magnitude of $\frac{\Delta H^\circ}{R}$ calculated from the above data is . . . . . . . (Note: The slope $m = -\frac{\Delta H^\circ}{2.303 R}$)
| $1/T \text{ (K}^{-1})$ | $\log_{10} K_p$ |
| $0.05$ | $3.5$ |
| $0.06$ | $2.5$ |
| $0.07$ | $1.5$ |
The magnitude of $\frac{\Delta H^\circ}{R}$ calculated from the above data is . . . . . . . (Note: The slope $m = -\frac{\Delta H^\circ}{2.303 R}$)
DifficultJEE MAIN 2026
View SolutionFor the reaction in equilibrium
$2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2(g)}$
if $P_{Br_2}$ is $\frac{P}{9}$ at equilibrium and $P$ is the total pressure,then calculate $\frac{P}{K_P}$.
$2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2(g)}$
if $P_{Br_2}$ is $\frac{P}{9}$ at equilibrium and $P$ is the total pressure,then calculate $\frac{P}{K_P}$.
Medium
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