For the following reaction $Fe_2O_{3(s)} + 3 CO_{(g)} \longrightarrow 2 Fe_{(s)} + 3 CO_{2(g)}$,$\Delta H^{\circ} = -29.8 \ kJ$ and $\Delta S^{\circ} = 15 \ JK^{-1}$. What is the value of $\Delta S_{\text{total}}$ at $298 \ K$ (in $JK^{-1}$)?

  • A
    $29.8$
  • B
    $100.0$
  • C
    $298.0$
  • D
    $115.0$

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$0.3 \ g$ of ethane undergoes combustion at $27^{\circ} C$ in a bomb calorimeter. The temperature of the calorimeter system (including the water) is found to rise by $0.5^{\circ} C$. The heat evolved during combustion of ethane at constant pressure is $....... kJ \ mol^{-1}$. (Nearest integer) [Given: The heat capacity of the calorimeter system is $20 \ kJ \ K^{-1}$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$. Assume ideal gas behaviour. Atomic mass of $C$ and $H$ are $12$ and $1 \ g \ mol^{-1}$ respectively]

The reaction $NH_2CN_{(s)} + \frac{3}{2}O_{2(g)} \to N_{2(g)} + CO_{2(g)} + H_2O_{(l)}$ is carried out in a bomb calorimeter. The heat evolved is $743 \ kJ \ mol^{-1}$. Calculate the value of $\Delta H$ at $300 \ K$ in $kJ \ mol^{-1}$.

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The enthalpies of combustion of $S_{(s)}$ and $H_{2(g)}$ are $-300 \ kcal \ mol^{-1}$ and $-290 \ kcal \ mol^{-1}$ respectively. Given the following reactions:
$SO_{3(g)} + H_2O_{(l)} \rightarrow H_2SO_{4(l)}$; $\Delta H = -130 \ kcal \ mol^{-1}$
$SO_{2(g)} + 1/2 O_{2(g)} \rightarrow SO_{3(g)}$; $\Delta H = -100 \ kcal \ mol^{-1}$
$S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}$; $\Delta H = -300 \ kcal \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)}$; $\Delta H = -290 \ kcal \ mol^{-1}$
The enthalpy of formation of $H_2SO_{4(l)}$ is:

Consider the following reaction:
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)} \to 6 CO_{2(g)} + 3 H_2O_{(g)}$
What are the signs of $\Delta H$,$\Delta S$,and $\Delta G$ for the above reaction?

$1 \, mol$ of oxygen gas is heated at constant volume from $20 \, ^oC$ to $30 \, ^oC$. What will be the change in the internal energy of the gas? Given the molar heat capacity of oxygen at constant pressure,$C_P = 7.03 \, cal \, mol^{-1} \, K^{-1}$ and $R = 2 \, cal \, mol^{-1} \, K^{-1}$. (Answer in $cal$)

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