Soru
h) Combustion of methanol may be represented by the following equation. 2CH_(3)OH_(4)(l)+3O_(2)(g)arrow 2CO_(2)(g)+4H_(2)O(l) Determine the molar enthalpy of combustion of methanol using the data given below C(s)+O_(2)(g)arrow CO_(2)(g) Delta H_(c)=-386.5kjmol^-1 H_(2)(g)+1/2O_(2)(g)arrow H_(2)O(l) Delta H_(f)=-264.5kjmol^-1 2C(s)+1/2O_(2)(g)+3H_(2)(g)arrow CH_(3)OH(l) Delta H_(f)=-256kjmol^-1 (9 marks)
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Elit · 8 yıl öğretmeniUzman doğrulaması
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To determine the molar enthalpy of combustion of methanol, we need to use Hess's Law, which states that the total enthalpy change of a reaction is the same regardless of the number of steps in which the reaction occurs. We can use the given equations to find the enthalpy change for the combustion of methanol.<br /><br />The balanced equation for the combustion of methanol is:<br />$2CH_{3}OH_{4}(l)+3O_{2}(g)\rightarrow 2CO_{2}(g)+4H_{2}O(l)$<br /><br />We can use the given equations to find the enthalpy change for each step of the reaction and then add them together to find the total enthalpy change.<br /><br />First, let's find the enthalpy change for the combustion of carbon:<br />$C(s)+O_{2}(g)\rightarrow CO_{2}(g)$<br />$\Delta H_{c}=-386.5kjmol^{-1}$<br /><br />Since there are two moles of carbon in the balanced equation, we need to multiply the enthalpy change by 2:<br />$2C(s)+2O_{2}(g)\rightarrow 2CO_{2}(g)$<br />$\Delta H_{c}=-773.0kjmol^{-1}$<br /><br />Next, let's find the enthalpy change for the combustion of hydrogen:<br />$H_{2}(g)+1/2O_{2}(g)\rightarrow H_{2}O(l)$<br />$\Delta H_{f}=-264.5kjmol^{-1}$<br /><br />Since there are four moles of hydrogen in the balanced equation, we need to multiply the enthalpy change by 4:<br />$4H_{2}(g)+2O_{2}(g)\rightarrow 4H_{2}O(l)$<br />$\Delta H_{f}=-1058.0kjmol^{-1}$<br /><br />Finally, let's find the enthalpy change for the formation of methanol:<br />$2C(s)+1/2O_{2}(g)+3H_{2}(g)\rightarrow 2CH_{3}OH(l)$<br />$\Delta H_{f}=-256kjmol^{-1}$<br /><br />Since there are two moles of methanol in the balanced equation, we need to multiply the enthalpy change by 2:<br />$2C(s)+1/2O_{2}(g)+3H_{2}(g)\rightarrow 2CH_{3}OH(l)$<br />$\Delta H_{f}=-512.0kjmol^{-1}$<br /><br />Now, we can add up the enthalpy changes for each step to find the total enthalpy change for the combustion of methanol:<br />$\Delta H_{c}+\Delta H_{f}+\Delta H_{f}=-773.0kjmol^{-1}-1058.0kjmol^{-1}-512.0kjmol^{-1}=-2443.0kjmol^{-1}$<br /><br />Therefore, the molar enthalpy of combustion of methanol is -2443.0 kJ/mol.
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