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How to start?? Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔrH∘A=+2035 kJ mol−1 2B(s)+3H2(g)→B2H6(g), ΔrH∘B=+36 kJ mol−1 H2(g)+12O2(g)→H2O(l), ΔrH∘C=−285 kJ mol−1 H2O(l)→H2O(g), ΔrH∘D=+44 kJ mol−1
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arrange the equations so the reactants and or the product are in the same side as in the reaction that you want to find the enthalpy for
for example the first reaction has to be reverted, with the B2O3 in the products side
cuanchi help me please
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