coal gas can undergo a progress called methanation 3H2(g) + CO(g)=CH4(g) +H2O(g) delta H=? Determine the enthalpy change involved in the reaction of 300g of carbon monoxide in this methanation reaction, using the following reference equation and enthalpy change   1) 2H2(g) + O2(g)= 2H2O(g) delta H1= - 483

ΔH = -2.20 x 10³ kJ

Step-by-step explanation

HESS' LAW:

• The enthalpies of discrete chemical processes can be used to determine the enthalpy of an overall reaction
• Our goal to solve for the enthalpy change for the methanation reaction is to modify the chemical equations and combine them to arrive with: ?3H2?(g)+CO(g)→CH4?(g)+H2?O(g)?
• Take note, modifying the chemical equation changes the value of ΔH

SOLUTION:

Modifying the chemical equations, we get.

?3H2?(g)+23?O2?(g)→3H2?O(g)ΔH1?=−725.4 kJ?

• The equation was multiplied by ?23??
• The value of ΔH is multiplied by ?23??

?CO(g)→C(s)+21?O2(g) ΔH2?=+110.5 kJ?

• The equation was reversed and divided by 2
• The value of ΔH is multiplied by 2 and its sign is reversed

?CO2?(g)+2H2?O(g)→CH4?(g)+2O2?(g) ΔH3?=+802.7 kJ?

• The equation was reversed
• The sign of ΔH is reversed

?C(s)+O2?(g)→CO2?(g)ΔH4?=−393.5 kJ?

• No changes were made

Thus, we arrive with the overall ΔH value below

?ΔHoverall?=−725.4 kJ+110.5 kJ+802.7 kJ−393.5 kJ=−205.7 kJ?

FOR THE REACTION OF 300 g of CO:

The amount of heat involved in the reaction of CO is determined below:

?300 g CO×28 g1 mol CO?×1 mol CO−205.7 kJ?=−2.20×103 kJ??