Answer to Question #157633 in General Chemistry for althea

Question

Calculate the ΔHo for methane gas , CH4, produced from graphite and hydrogen gas according to the following reaction: C(graphite) + 2H2(g) à CH4(g)


Data given:

C(graphite) + O2(g) à CO2(g) ΔHo= -393.5 kJ
H2(g) + ½ O2(g) à H2O(l) ΔHo= - 285.8 kJ
CH4(g) + 2O2(g) à CO2(g) + 2H2O(l) ΔHo = -890.4kJ

Answer 1

C_(graphite) + O_2(g) $\to$ CO_2(g) ΔH^o= -393.5 kJ ...................................(i)

H_2(g) + ½ O_2(g)$\to$ H₂O_(l) ΔH^o= - 285.8 kJ ........................................(ii)

CH_4(g) + 2O_2(g) $\to$ CO_2(g) + 2H₂O_(l) ΔHo = -890.4kJ .....................(iii)

We need to use the equations above to Calculate the ΔHo for methane gas, CH4, produced from graphite and hydrogen gas.

C_(graphite) + 2H_2(g) $\to$ CH_4(g)

First, we need to rearrange the equations and balance them out so that the same substances can cancel out.

We will then reverse equation (iii) and multiply equation (ii) by 2 as shown;

C_(graphite) + O_2(g) $\to$ CO_2(g) ΔH^o= -393.5 kJ ..........................................(i)

2H_2(g) + O_2(g)$\to$ 2H₂O_(l) ΔH^o= - 571.6 kJ ........................................(ii)

CO_2(g) + 2H₂O_(l) $\to$ CH_4(g) + 2O_2(g) ΔHo = 890.4kJ .............................(iii)

If we cross out all the highlighted substances in the equations we remain with the equation;

C_(graphite) + 2H_2(g) $\to$ CH_4(g)

Thus the ΔH^o will be added as follows;

ΔH^o = (-393.5) + (-571.6) + (890.4) kJ

= (-965.1) + (890.4) kJ

= -74.7kJ