Methanol is a clean-burning liquid fuel being considered as a replacement for ga

Question

Methanol is a clean-burning liquid fuel being considered as a replacement for gasoline. It can be produced by the controlled reaction of the oxygen in air with methane in natural gas (reaction (1) CH_4(g) + 1/2 O_2(g) rightarrow CH_3OH_(l) Use only the following information to calculate the Delta H_f degree for this reaction: CH_4(g) + H_2O_(g) rightarrow CO_(g) + 3 H_2(g) Delta H_r degree = 206.10 kJ 2 H_2(g) + CO_(g) rightarrow CH_3OH_(l) Delta H_f degree = -128.33 kJ Delta H_f degree (H_2O) = -241.82 kJ

Explanation / Answer

Q3.

For CH4 + 1/2O2 = CH3OH

We need CH4 in the left so keep equation (2) as it is...

We need CH3OH in the right, so keep Eqn 3 as it is

so

add both equation

CH4 + H2O + 2H2 + CO = CO + 3H2 + CH3OH HRxn = 206.10 - 128.33 = 77.77 kJ/mol

CH4 + H2O = H2 + CH3OH HRxn = 77.77 kJ/mol

we know that H fo rH2O = -241.82 and H fo rH2 = 0 due to elemental so

CH4 + -241.82 = 0 + CH3OH HRxn = 77.77 kJ/mol

Then

Hrreaction = 77.77+241.82 = 319.59 kJ/mol

b)

Use:

HRxn = Hproducts - Hreactants

HRx = -238.4 - (-74.87 + 1/2*0) = -163.53 kJ/mol

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