Here is a problem related to Hess's Law for the student to solve:Suppose you are given the following three equations, each with their respective enthalpy change: (1) 2H2(g) + O2(g) --> 2H2O(l) ΔH₁ = -572 kJ/mol (2) 2H2O(l) --> 2H2(g) + O2(g) ΔH₂ = 472 kJ/mol (3) C(s) + O2(g) --> CO2(g) ΔH₃ = -394 kJ/molUse Hess's law to calculate the enthalpy change (in kJ/mol) for the following reaction: C(s) + 2H2(g) --> CH4(g)

Question

Here is a problem related to Hess's Law for the student to solve:Suppose you are given the following three equations, each with their respective enthalpy change:     (1) 2H2(g) + O2(g) --> 2H2O(l)   ΔH₁ = -572 kJ/mol    (2) 2H2O(l) --> 2H2(g) + O2(g)   ΔH₂ = 472 kJ/mol    (3) C(s) + O2(g) --> CO2(g)   ΔH₃ = -394 kJ/molUse Hess's law to calculate the enthalpy change

Answer 1

To use Hess's law, we need to manipulate the given equations so that they can be combined to form the target equation: C s  + 2H2 g  --> CH4 g . First, we need to reverse equation  2  to get H2 g  and O2 g  on the reactant side: 2'  2H2 g  + O2 g  --> 2H2O l    H' = -472 kJ/molNow, we need to form CH4 g  in the products. We can do this by adding the following equation: 4  C s  + 2H2O l  --> CH4 g  + O2 g    H = x kJ/molWe can obtain equation  4  by combining equations  1 ,  2' , and  3 . Let's do that: 1  +  2'  +  3  =  4 2H2 g  + O2 g  --> 2H2O l    H = -572 kJ/mol2H2 g  + O2 g  --> 2H2O l    H' = -472 kJ/molC s  + O2 g  --> CO2 g    H = -394 kJ/mol-----------------------------------------C s  + 2H2O l  + 2H2 g  + O2 g  --> 2H2O l  + CO2 g  + CH4 g  + O2 g Simplifying the equation, we get:C s  + 2H2 g  --> CH4 g  + CO2 g Now, we need to subtract equation  3  from the combined equation to get the target equation: 4  -  3  = Target EquationC s  + 2H2 g  --> CH4 g  + CO2 g    H = x kJ/mol-  C s  + O2 g  --> CO2 g    H = -394 kJ/mol -----------------------------------------C s  + 2H2 g  --> CH4 g    H = x + 394 kJ/molNow, we can find the enthalpy change for the target equation by adding the enthalpy changes of the combined equations:H = H + H' + HH = -572 kJ/mol +  -472 kJ/mol  +  -394 kJ/mol H = -1438 kJ/molSo, the enthalpy change for the target equation is:H = x + 394 kJ/molx = -1438 kJ/mol - 394 kJ/molx = -1832 kJ/molTherefore, the enthalpy change for the reaction C s  + 2H2 g  --> CH4 g  is -1832 kJ/mol.