MHT CET 2026 13th April Morning Shift | Thermodynamics Question 40 | Chemistry

Calculate the enthalpy change of the reaction,
$\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)}$
if bond energies (kJ mol$^{-1}$):
H–H = 436, Cl–Cl = 242, H–Cl = 431

$-184$ kJ/mol

$-246$ kJ/mol

$-242$ kJ/mol

$-431$ kJ/mol

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

The equilibrium constant for a reaction is $100$. What will be the value of standard Gibbs energy change at $298$ K ? ($R = 8.314$ J $\text{K}^{-1}\text{mol}^{-1}$)

$-11.411$ KJ/mol

$-5.744$ KJ/mol

$-570.584$ KJ/mol

$-57.058$ KJ/mol

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

The bond dissociation enthalpy of $\text{H}_2$, $\text{Cl}_2$ and HCl are $434$, $242$ and $431$ kJ $\text{mol}^{-1}$ respectively. Calculate the enthalpy of formation of HCl.

$-93$ kJ $\text{mol}^{-1}$

$245$ kJ $\text{mol}^{-1}$

$93$ kJ $\text{mol}^{-1}$

$-245$ kJ $\text{mol}^{-1}$

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