What is the ∆g for the following reaction under standard conditions (t = 298 k) for the formation of nh4no3(s)? 2nh3(g) + 2o2(g) nh4no3(s) + h2o(l)

given:
nh4no3(s): ∆hf = -365.56 kj ∆sf = 151.08 j/k.
nh3(g): ∆hf = -46.11 kj ∆sf = 192.45 j/k.
h2o(l): ∆hf = -285.830 kj ∆sf = 69.91 j/k.
o2(g): ∆hf = 0.00 kj ∆sf = 205 j/k.

is it a purebred? it looks more like a red nose, but it doesn't have the red could it be a staffordshire bull terrior? unsure if that's the right name for it sorry

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol