If $\Delta_r H^{\ominus}$ and $\Delta_r S^{\ominus}$ are standard enthalpy change and standard entropy change respectively for a reaction,the incorrect option is

  • A
    $\Delta_r H^{\ominus} = \text{negative}; \Delta_r S^{\ominus} = \text{positive}; \text{spontaneous at all temperatures}$
  • B
    $\Delta_r H^{\ominus} = \text{negative}; \Delta_r S^{\ominus} = \text{negative}; \text{non-spontaneous at low temperatures}$
  • C
    $\Delta_r H^{\ominus} = \text{positive}; \Delta_r S^{\ominus} = \text{positive}; \text{non-spontaneous at low temperatures}$
  • D
    $\Delta_r H^{\ominus} = \text{negative}; \Delta_r S^{\ominus} = \text{negative}; \text{spontaneous at low temperatures}$

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For a certain reaction, $\Delta H = -50 \text{ kJ}$ and $\Delta S = -100 \text{ J/K}$. Find the temperature range at which the reaction is spontaneous.

The standard state Gibbs free energies of formation of $C$ (graphite) and $C$ (diamond) at $T = 298 \ K$ are:
$\Delta_f G^0[C(\text{graphite})] = 0 \ kJ \ mol^{-1}$
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The standard state means that the pressure should be $1 \ bar$,and the substance should be pure at a given temperature. The conversion of graphite [$C$ (graphite)] to diamond [$C$ (diamond)] reduces its volume by $2 \times 10^{-6} \ m^3 \ mol^{-1}$. If $C$ (graphite) is converted to $C$ (diamond) isothermally at $T = 298 \ K$,the pressure at which $C$ (graphite) is in equilibrium with $C$ (diamond) is:
[Useful information: $1 \ J = 1 \ kg \ m^2 \ s^{-2} ; 1 \ Pa = 1 \ kg \ m^{-1} \ s^{-2} ; 1 \ bar = 10^5 \ Pa$ ] (in $bar$)

For $2 \ mol$ of $He$ (ideal gas),find the work done in a process where it is heated from $200 \ K$ to $400 \ K$ such that the enthalpy of the gas varies as : $H = 10 \ V^2$ (in $R$)

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The relation between $\Delta G$ and $\Delta H$ is

Which of the following conditions regarding a chemical process ensures its spontaneity at all temperatures?

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