Question

Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) => CH 3OH(l) Using the following information: C(graphite) + O 2 => CO 2(g) ΔH o = -393.5 kJ H 2(g) + 1/2 O 2 => H 2O(l) ΔH o = -285.8 kJ CH 3OH (l) + 3/2 O 2(g) => CO 2(g) + 2H 2O(l) ΔH o = -726.4 kJ a. +238.7 kJ b. -238.7 kJ c. +548.3 kJ d. -548.3 kJ e. +904.5 kJ

Answer 1

Answer:

$\Delta H$ for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

$C(graphite)+O_{2}(g)\rightarrow CO_{2}(g)$ $\Delta H_{1}= -393.5 kJ$

$2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)$ $\Delta H_{2}= (2\times -285.8)kJ$

$CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)$  $\Delta H_{3}= 726.4 kJ$

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$C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)$

$\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ$