Assertion : $\Delta H$ and $\Delta E$ are almost same for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$.
Reason : All reactants and products are gases.
Reason : All reactants and products are gases.
- AIf both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
- BIf both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
- CIf the Assertion is correct but Reason is incorrect.
- DIf both the Assertion and Reason are incorrect.
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Similar Questions
At $300 \ K$, $3.0 \ \text{moles}$ of an ideal gas at $3.0 \ \text{atm}$ pressure is compressed isothermally to one half of its volume by an external pressure of $6.0 \ \text{atm}$. The work done (in $kJ$) is. Given, $R=0.082 \ \text{L atm K}^{-1} \text{mol}^{-1}$ $(1 \ \text{L atm} = 101.3 \ \text{J})$.
Calculate $\Delta H^{\circ}$ for the reaction, $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$, given the following reactions:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2}H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2}H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$
DifficultTS EAMCET 2009
View SolutionStandard entropies of $X_2$,$Y_2$ and $XY_3$ are $60$,$40$ and $50 \ J \ K^{-1} \ mol^{-1}$ respectively. For the reaction $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \rightarrow XY_3$,the enthalpy change is $\Delta H = -30 \ kJ \ mol^{-1}$. At what temperature will the reaction be at equilibrium (in $K$)?
$CsOH + HCl \to CsCl + H_2O$,$\Delta H = -13.4 \ K \ cal/mol$
$CsOH + HA \to CsA + H_2O$,$\Delta H = -10.4 \ K \ cal/mol$
Then calculate $\Delta H$ of ionisation of $HA$ in $K \ cal/mol$.
$CsOH + HA \to CsA + H_2O$,$\Delta H = -10.4 \ K \ cal/mol$
Then calculate $\Delta H$ of ionisation of $HA$ in $K \ cal/mol$.
Medium
View SolutionCalculate the enthalpy change when $6 \text{ g}$ of $CO(g)$ reacts with sufficient $NO_2(g)$ according to the following reaction: $4 \text{ CO}(g) + 2 \text{ NO}_2(g) \rightarrow 4 \text{ CO}_2(g) + 2 \text{ N}_2(g)$; $\Delta_r H^0 = -1200 \text{ kJ}$ (in $\text{ kJ}$)
DifficultMHT CET 2026
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