For a reaction H_{2(g)} + frac{1}{2}O_{2(g)} | vedclass

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(B) Using Kirchhoff's equation: $\Delta H_2 = \Delta H_1 + \Delta C_p(T_2 - T_1)$.Given: $\Delta H_1 = -285.8 \ kJ \ mol^{-1} = -285800 \ J \ mol^{-1}

Vedclass · Accepted Answer

$-282.6$

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(B) Using Kirchhoff's equation: $\Delta H_2 = \Delta H_1 + \Delta C_p(T_2 - T_1)$.Given: $\Delta H_1 = -285.8 \ kJ \ mol^{-1} = -285800 \ J \ mol^{-1}$,$\Delta C_p = 32 \ J \ K^{-1}$,$T_1 = 27 + 273 = 300 \ K$,$T_2 = 127 + 273 = 400 \ K$.$\Delta H_2 = -285800 + 32(400 - 300)$.$\Delta H_2 = -285800 + 32(100) = -285800 + 3200 = -282600 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$: $\Delta H_2 = -282.6 \ kJ \ mol^{-1}$.

For a reaction $H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta C_p = 32 \ J \ K^{-1}$,$\Delta H$ at $27 \ ^oC = -285.8 \ kJ \ mol^{-1}$. What will be the value of $\Delta H$ at $127 \ ^oC$ in $kJ \ mol^{-1}$?

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