Question

Calculate enthalpy change for the reaction at 298 K
NH4Cl(aq)+NaNO2(S)---- → NaCl(s)+2H2O(L)+N2(g)
Internal energy change -332kj​

Answer 1

Answer:

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Answer 2

Given: $NH_{4} Cl+NaNO_{2} -->NaCl+2H_{2} O+N_{2}$. Δ$U=-332kJ$

To find: enthalpy change at 298K.

Step-by-step explanation:

Step 1 of 2

The enthalpy change is calculated by the formula: ΔH=ΔU+ΔngRT.

$NH_{4} Cl+NaNO_{2} -->NaCl+2H_{2} O+N_{2}$

Here Δ$n_{g} =(1+2+1)-(1+1)=2$

Step 2 of 2

ΔH=ΔU+ΔngRT

substituting all the values. The enthalpy change will be:

$=(-332)kJ+(2*8.314*298)\\=-332kJ+4955.144J\\=-332kJ+4.955kJ\\=-327.045kJ$

The enthalpy change is -327.045kJ.