The combination of coke and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. If we assume coke can be represented by graphite, the equation for the production of coal gas is

2C(s)+2H2O(g)→CH4(g)+CO2(g)

a.) Determine the standard enthalpy change for this reaction from the following standard enthalpies of reaction:

i. C(s)+H2O(g)→CO(g)+H2(g):ΔH0=131.3kJ

ii. CO(g)+H2O(g)→CO2(g)+H2(g):ΔH0=−41.2 kJ

iii. CH4(g)+H2O(g)→3H2(g)+CO(g):ΔH0=206. 1kJ

15.3kJ

Explanation:

Hello,

This problem could be solved by the Hess Law which allows us modify the chemical reactions and their directions to find a desired one as follows:

- The first step is to multiply the first reaction by 2 in order to get 2 moles of C and H2O:

2(C(s)+H2O(g)→CO(g)+H2(g):ΔH0=131.3kJ)=2C(s)+2H2O(g)→2CO(g)+2H2(g);ΔH0=262.6kJ

- Next, the third reaction is inverted:

CH4(g)+H2O(g)→3H2(g)+CO(g):ΔH0=206.1kJ ... 3H2(g)+CO(g)→CH4(g)+H2O(g):ΔH0=-206.1kJ

- Then, the second reaction remains the same.

CO(g)+H2O(g)→CO2(g)+H2(g):ΔH0=−41.2kJ

- Finally, the three reactions are added in order to obtain:

2C(s)+2H2O(g)+3H2(g)+CO(g)+CO(g)+H2O(g)→2CO(g)+2H2(g)+CH4(g)+H2O(g)+CO2(g)+H2(g)

So just the bold terms remain to give the desired reaction:

2C(s)+2H2O(g)→CH4(g)+CO2(g)

Thus, the standard enthalpy change result:

ΔH0=262.6kJ-206.1kJ−41.2kJ=15.3kJ

Best regards.

hi my dude or dudette