Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per the following endothermic reaction:
$CH_{4(g)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + 3H_{2(g)}$
$(a)$ Write an expression for $K_p$ for the above reaction.
$(b)$ How will the values of $K_p$ and the composition of the equilibrium mixture be affected by:
$(i)$ increasing the pressure
$(ii)$ increasing the temperature
$(iii)$ using a catalyst?

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(A) The expression for the equilibrium constant $K_p$ is:
$K_p = \frac{p_{CO} \times p_{H_2}^3}{p_{CH_4} \times p_{H_2O}}$
$(b)$ $(i)$ Increasing the pressure shifts the equilibrium in the backward direction because the number of moles of gaseous products $(4 \ mol)$ is greater than the number of moles of gaseous reactants $(2 \ mol)$. $K_p$ remains unchanged as it depends only on temperature.
$(ii)$ Increasing the temperature shifts the equilibrium in the forward direction because the reaction is endothermic $(\Delta H > 0)$. Consequently,the value of $K_p$ increases.
$(iii)$ Using a catalyst does not affect the equilibrium position or the value of $K_p$. It only increases the rate at which the equilibrium is attained.

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At $600 \ K$,$2 \ mol$ of $NO$ are mixed with $1 \ mol$ of $O_2$.
$2 \ NO_{(g)} + O_{2(g)} \rightleftarrows 2 \ NO_{2(g)}$
The reaction occurring as above comes to equilibrium under a total pressure of $1 \ atm$. Analysis of the system shows that $0.6 \ mol$ of oxygen are present at equilibrium. The equilibrium constant for the reaction is $.........$ (Nearest integer).

For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,what is the relationship between $\Delta H$ and $\Delta E$?

Calculate:
$(a)$ $\Delta G^{\circ}$ and
$(b)$ the equilibrium constant for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \, K$
$NO_{(g)} + 1/2 O_{2(g)} \longleftrightarrow NO_{2(g)}$
Given:
$\Delta G^{\circ}_f(NO_2) = 52.0 \, kJ/mol$
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When $0.087 \ mol$ of $NO$ and $0.0437 \ mol$ of $Br_{2}$ are mixed in a closed container at constant temperature,$0.0518 \ mol$ of $NOBr$ is obtained at equilibrium. Calculate the equilibrium amount of $NO$ and $Br_{2}$.

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
$CH_3COOH_{(l)} + C_2H_5OH_{(l)} \longleftrightarrow CH_3COOC_2H_{5(l)} + H_2O_{(l)}$
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