For the complete combustion of ethene,$C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}$,the amount of heat produced as measured in a bomb calorimeter is $1406 \ kJ \ mol^{-1}$ at $300 \ K$. The minimum value of $T \Delta S$ needed to reach equilibrium is $(-)....... \ kJ$. (Nearest integer) Given: $R = 8.3 \ J \ K^{-1} \ mol^{-1}$

  • A
    $1411$
  • B
    $1412$
  • C
    $1413$
  • D
    $1414$

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Similar Questions

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27\,^oC$. If the work done during the process is $3\,kJ$,the final temperature will be equal to $(C_v = 20\,J\,K^{-1} \, mol^{-1})$

Calculate the work done during the combustion of $0.138 \ kg$ of ethanol,$(C_2H_5OH_{(l)})$ at $300 \ K$. Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$ and molar mass of ethanol $= 46 \ g \ mol^{-1}$. (in $J$)

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 $:-$

Consider the following data for the reaction $X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$ at $600 \ K$. The $\Delta_r G^\circ$ (in $kJ \ mol^{-1}$) for the reaction is:
Compound $\Delta_f H^\circ$ $(kJ \ mol^{-1})$ $S^\circ$ $(J \ mol^{-1} \ K^{-1})$
$XY(g)$ $42$ $200$
$X_2(g)$ $8$ $140$
$Y_2(g)$ $80$ $250$

Match the following:
$(a)$ Entropy of vaporization $(1)$ Decreases
$(b)$ $K$ for spontaneous process $(2)$ Always has a $(+)$ value
$(c)$ Crystalline solid state $(3)$ Has minimum entropy
$(d)$ $\Delta U$ for adiabatic expansion of an ideal gas $(4)$ $\frac{\Delta H_{vap}}{T_b}$

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