Q.For the reaction , kJ and JK. Calculate for the reaction, and predict whether the reaction may occur spontaneously.
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Start your 14-day free trial to unlock the full solution →Using , we first find from via , then compute at 298 K. The result is positive, so the reaction is non-spontaneous at this temperature.
The key to solving this lies in connecting two thermodynamic quantities: internal energy change () and enthalpy change (), and then using Gibbs free energy to judge spontaneity. You're given and , but the Gibbs equation uses , not . So the first step is always to convert.
Why? Because is measured at constant volume, while most reactions (including this one) occur at constant pressure (open container). The enthalpy change accounts for the pressure-volume work done by or on the system. The relation is:
where is the change in moles of gas.
Let's work through it step by step.
-
Find
For the reaction :
Moles of gaseous products = 2
Moles of gaseous reactants = 2 + 1 = 3
So .
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Calculate
Given kJ = J (we'll work in J for consistency with in J/K).
Use J mol K and assume standard temperature K (since not specified, this is the default for such problems).
Compute J.
So J kJ.
The negative tells us the reaction is exothermic — it releases heat. But that alone doesn't guarantee spontaneity; entropy also matters.
- Apply the Gibbs free energy equation
Given J K.
At K:
Compute J.
So:
That's about kJ. …
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