Thermodynamics of Equilibrium

The Haber’s process for the formation of ${\mathrm{NH}}_3$ at $298 \mathrm{K}$ is ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3;\mathrm{\hspace{0.17em}}\mathrm{\Delta H}=-46.0\mathrm{KJ}$. Which of the following is the correct statement?

A schematic plot of ln  $K_{\mathrm{eq}}$ versus inverse of temperature for a reaction is shown below

The reaction must be –

In a one-litre flask, $6$ moles of $\mathrm{A}$ undergoes the reaction $\mathrm{A}( \mathrm{g})\rightleftharpoons \mathrm{P}( \mathrm{g})$. The progress of product formation at two temperatures (in Kelvin), ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$, is shown in the figure:

If ${\mathrm{T}}_1=2{\mathrm{T}}_2$ and$\left({\mathrm{\Delta G}}_2^{\mathrm{o}}-{\mathrm{\Delta G}}_1^{\mathrm{o}}\right)={\mathrm{RT}}_2\mathrm{lnx}$ , then the value of $\mathrm{x}$ is
[ ${\mathrm{\Delta G}}_1^{\mathrm{o}}$ and ${\mathrm{\Delta G}}_2^{\mathrm{o}}$ are standard Gibb's free energy change for the reaction at temperatures ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$, respectively.]

What is the (approx) partial pressure of ${\mathrm{CO}}_2$at equilibrium at $427°\mathrm{C}$?

${\mathrm{CaCO}}_{3\left(\mathrm{s}\right)}\stackrel{}{\rightleftharpoons }{\mathrm{CaO}}_{\left(\mathrm{s}\right)}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)} ∆{\mathrm{G}}^{\circ }=-13.5 \mathrm{kJ}$

Select the correct option about $\mathrm{\Delta G}$ for the following reaction

$\begin{array}{ccc}\underset{(2 \mathrm{atm},373 \mathrm{K})}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}& ⟶& \underset{ (2 \mathrm{atm},373 \mathrm{K})}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}\end{array}$

(Given: $0.0821\times 373\times \ln 2=21.23 \mathrm{L} \mathrm{atm}$, molar volume of water at $2 \mathrm{atm}, 373 \mathrm{K}=18 \mathrm{ml}$)

$\mathrm{\Delta }{\mathrm{G}}^{\mathrm{o}}=+\mathrm{ }63.3\mathrm{ }\mathrm{k}\mathrm{J}$ for the reaction ${\mathrm{Ag}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)\to 2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CO}}_3^{-2}\left(\mathrm{aq}\right)$
the thermodynamic equilibrium constant at ${25}^{\mathrm{o}}\mathrm{C}$ is:

Which of the following is correct at equilibrium?

Select the graph which can be used to determine standard free energy of the reaction given below:

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$
Assume that the reaction is at equilibrium.

Among the following set of conditions, which condition necessarily holds true for a non-feasible process?
(Where, $\mathrm{K}=$equilibrium constant)

Calculate the temperature given that the value of ${\mathrm{K}}_{\mathrm{c}} = {10}^3$ and ${\mathrm{\Delta G}}_{\mathrm{f}}°$of $\mathrm{A}, \mathrm{B} \mathrm{and} \mathrm{C}$ are $-100, -200 \mathrm{and} -500 \mathrm{kJ}/\mathrm{mol}$and the reactants $\mathrm{A}$ and $\mathrm{B}$ is in equilibrium with $\mathrm{C}$.

Calculate $∆\mathrm{G}$ for the following change, if vapour pressure of water is $0.01 \mathrm{atm}$ at $27º\mathrm{C}$.

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}, 0.01 \mathrm{atm}, 27°\mathrm{C}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}, 0.01 \mathrm{atm}, 27°\mathrm{C}\right)$

Which is correct for the combustion reaction occurring in an automobile is $2{\mathrm{C}}_8{\mathrm{H}}_{18}\left(\mathrm{s}\right) + 25{\mathrm{O}}_2\left(\mathrm{g}\right) \to 16{\mathrm{CO}}_2\left(\mathrm{g}\right) + 18{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$. 

The value of ${\mathrm{K}}_{\mathrm{c}}$ for the following reaction at $298 \mathrm{K}$ is:

$2{\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{NH}}_2{\mathrm{CONH}}_2\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{\ell }\right)$  

Given, standard Gibbs energy change $\left(\mathrm{\Delta G}°\right)$ at the given temperature is $-13.6\mathrm{kJ}/{\mathrm{mol}}^{-1}$

For the following reaction,

${\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$

the ${\mathrm{k}}_{\mathrm{p}}=0.05 \mathrm{atm}$. What is the value of $\mathrm{\Delta G}°$ of the reaction at ${627}^{\circ }\mathrm{C}$?

For a spontaneous chemical reaction, the value of $∆\mathrm{G}°$ must be negative and the value of ${\mathrm{K}}_{\mathrm{eq}}$ must be greater than __

Given: $∆\mathrm{G}° = -\mathrm{RT} \ln {\mathrm{K}}_{\mathrm{eq}}$

Consider that for a hypothetical reaction, $\mathrm{\Delta H}°$ and $\mathrm{\Delta S}°$ are $-30 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and $-100 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$ at $300\mathrm{K}.$ What is the equilibrium constant for the reaction at $298 \mathrm{K}?$

For a hypothetical reaction, $\Delta \mathrm{H}=-30 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and $\Delta \mathrm{S}=-100 J K^{-1} {\mathrm{mol}}^{-1}.$ At what temperature (in Celsius), the reaction is at equilibrium?

For the given reaction, concentrations are given as: $\left[{\mathrm{NH}}_3\right]$ is $0.05 \mathrm{M}$ and $\left[{{\mathrm{NH}}_4}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]=0.002 \mathrm{M}$ in the presence of excess water. What is the value of $\Delta \text{G}\left(\text{KJ}/\text{mol}\right)$ at $298 \mathrm{K}$ at some non-equilibrium condition?

$\Delta {\text{G}}_{\text{Reaction}}^{\circ }=+26·81\text{ KJ}\text{ mol}$
${\mathrm{NH}}_3\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟶{\mathrm{NH}}_4^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Consider the curve between $\ln \mathrm{K}$ and $1/\mathrm{T}$. Which plot will be correct for exothermic reaction?