ANSWER: Write the given reaction. 2 H2O(g) = 2 H2(g) + O2(g) deltaH = 483.6 kJ/m

Question

ANSWER:

Write the given reaction.
2 H2O(g) = 2 H2(g) + O2(g) deltaH = 483.6 kJ/mol
Temperature = 25 C or 323 K

Step 1: Calculate the delta n = (Sum of product moles - Sum of reactant moles).
= ((2+1)-2)= 1mol
Step 2.

Calculate the work done.

W = RT x Delta n
= (( 8.314 x10^-3 KJ/mol K) x (322 K) x1mol)
= 2.677 KJ

Step 3:

Calculate the internal energy changes of this reaction,
Formula: Delta U = Delta H - (RT x Delta n)
= (2 mol x 483.6 kJ/mol)- (( 8.314 x10^-3 KJ/mol K) x (322 K) x1mol)
= (967.2 KJ - 2.677 KJ)
= 964.523 KJ.
So the Internal Energy Changes of this reaction is 964.523 KJ

Explanation / Answer

please solve this question step by step.

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