Question

in problems 4 - 7, calculate δh for the reaction:
3ch4(g)→c3h8(g)+2h2(g)

1. using the thermochemical equations

2c3h8(g)+10o2(g)→6co2(g)+8h2o(ℓ)
δh=-4441kj
ch4(g)+2o2(g)→co2(g)+2h2o(ℓ)
δh=-890kj
2h2(g)+o2(g)→2h2o(ℓ)
δh=-572kj

Explanation:

Step1: Reverse the first equation

$$2H_2O(l)\to2H_2(g)+O_2(g)\quad\Delta H = + 572\space kJ$$

Step2: Multiply the second equation by 2

$$2CH_4(g)+4O_2(g)\to2CO_2(g)+4H_2O(l)\quad\Delta H=-1780\space kJ$$

Step3: Add the reversed first equation, the multiplied - second equation and the third equation

$$3CH_4(g)\to C_3H_8(g)+2H_2(g)$$
$$\Delta H=(+ 572)+(-1780)+(-4441)$$
$$\Delta H = 572-1780 - 4441$$
$$\Delta H=-5649\space kJ$$

Answer:

$\Delta H=-5649\space kJ$