For which of the following processes will the value of $\Delta H^{\circ} - \Delta G^{\circ}$ be approximately zero?

  • A
    $CaCO_{3(s)} \to CaO_{(s)} + CO_{2(g)}$
  • B
    $FeSO_{4(s)} + Zn_{(s)} \to ZnSO_{4(s)} + Fe_{(s)}$
  • C
    $Zn_{(s)} + H_2SO_{4(aq)} \to ZnSO_{4(s)} + H_{2(g)}$
  • D
    $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$

Explore More

Similar Questions

Two liters of $N_2$ gas at $0 \, ^\circ C$ and $5 \, atm$ pressure undergoes isothermal expansion against a constant external pressure of $1 \, atm$ until the pressure of the gas becomes $1 \, atm$. If the gas is ideal,what is the work done in the expansion in $J$?

Difficult
View Solution

For the reaction $2H_{(g)} \to H_{2(g)}$,the signs of $\Delta H$ and $\Delta S$ are:

Calculate the heat produced in $kJ$ when $280 \ g$ of $CaO$ is completely converted to $CaCO_3$ by reaction with $CO_2$ at $27 \ ^{\circ}C$ and at constant volume :-
(Given) $\Delta H^o_f (CaCO_3, s) = -1207 \ kJ/mol$
$\Delta H^o_f (CaO, s) = -635 \ kJ/mol$
$\Delta H^o_f (CO_2, g) = -394 \ kJ/mol$ (in $kJ$)

Difficult
View Solution

The value of $\Delta H_{transition}$ for $C(\text{graphite}) \rightarrow C(\text{diamond})$ is $1.9 \ kJ/mol$ at $25^{\circ}C$. The entropy of graphite is higher than the entropy of diamond. This implies that which of the following is incorrect $:-$

$5 \, mol$ of an ideal gas at $100 \, K$ are allowed to undergo reversible compression till its temperature becomes $200 \, K$. If $C_v = 28 \, J \, K^{-1} \, mol^{-1}$,calculate $\Delta U$ and $\Delta pV$ for this process. $(R = 8.0 \, J \, K^{-1} \, mol^{-1})$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo